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Analysis

We have $n$ knights labeled $0,1,\dots,n-1$. Knight 0 is the King (the root of the hierarchy). For each knight $i\in [1..n-1]$, we are given a direct superior $\text{arr}[i-1]$. This induces a tree (or forest, if you momentarily remove the King) where node $0$ is the root, and there is a parent $\text{arr}[i-1]\to i$.

A knight $X$ is a (direct or indirect) superior of knight $Y$ if and only if $X$ is on the unique path from the root down to $Y$. Equivalently, in tree terms:

• $X$ is an ancestor of $Y$ (possibly the same node if $X=Y$, but in this problem we only worry about the strict ancestor case).

We want to form squads (pairs) of two distinct knights $(u,v)$ such that:

1. Neither is superior to the other. In tree terms, $u$ is not an ancestor of $v$ and $v$ is not an ancestor of $u$. That is, there is no ancestor/descendant relationship between them.
2. Each knight can appear in at most one pair.

The goal is to maximize the number of such disjoint pairs.

A few key observations:

1. Knight 0 (the King) can never pair with anyone, because the King is an ancestor of all other knights. So effectively, only knights in $\{1,\dots,n-1\}$ can form pairs.

2. If the tree is a chain (like $0\to 1\to 2\to \dots$), then every node except the root is an ancestor or descendant of every other node in that chain. Hence no pairs can be formed in a pure chain (the answer would be 0).

3. In a star shape (root $0$ has children $1,2,\dots,n-1$), then none of the non-root nodes are ancestors of each other—so we can freely pair them up. We get $\lfloor \tfrac{n-1}{2}\rfloor$ pairs.

4. In general, we have a tree of size $n$. Removing the root (knight 0) leaves us with a forest of subtrees. However, knights inside the same subtree can sometimes still form valid pairs (e.g., two siblings). Knights across different subtrees of the root also can form pairs because they do not lie on each other’s ancestor–descendant chain. In fact, in more complex hierarchies, one can form cross-pairs deeper in the tree (as in the sample example).

5. Formally, if we look at the “incomparability graph” $G$ on vertices $\{1,\dots,n-1\}$ with an edge $(u,v)$ whenever $u$ is not an ancestor of $v$ and $v$ is not an ancestor of $u$, then the problem is to find a maximum matching in $G$. This graph is the incomparability graph of the tree’s partial order (“ancestor” is the partial order). Such graphs (the incomparability graphs of a tree poset) are known to be cographs, and there exist linear/near-linear time algorithms for maximum matching in cographs via modular decomposition. However, implementing the cograph-matching algorithm from scratch can be non-trivial.

6. If the problem constraints are small (say $n \leq 2000$ or so), one can do the following simpler (but heavier) approach:

   • Perform a DFS to get entry/exit time for each node: $\text{in}[u]$, $\text{out}[u]$. Then $$u \text{ is ancestor of } v \iff \text{in}[u] \le \text{in}[v] \;\text{and}\; \text{out}[u] \ge \text{out}[v].$$
   • Build the graph among knights $\{1,\dots,n-1\}$. For $u < v$, add an edge $(u,v)$ if neither is ancestor of the other.
   • Run a maximum matching algorithm in a general graph (e.g. Edmonds’ Blossom algorithm).

   This is $O(n^2)$ to build the graph plus $O(n^3)$ in a naive Blossom implementation. That might be acceptable for $n$ up to a few thousands, but not for very large $n$.

7. If the problem constraints are large (tens or hundreds of thousands), one must implement a more specialized cograph matching or use a known decomposition approach in near-linear time. Because the problem statement does not provide explicit constraints, below we’ll show the more straightforward “build adjacency + blossom” approach in C++ (assuming a moderate $n$). If you need to handle very large $n$, you’d pursue the cograph/modular-decomposition method.

Below is a reference-style C++ code that:

1. Reads the input array arr[] of length $n-1$.
2. Builds an adjacency list for the tree of size $n$.
3. Runs a DFS from the root (0) to compute inTime[u] and outTime[u].
4. Builds the “incomparability” graph among $\{1,\dots,n-1\}$.
5. Runs a standard “Edmonds Blossom” algorithm to find the maximum matching.
6. Prints the size of that matching.

Again, important disclaimer: if $n$ is large (e.g., $10^5$), the $O(n^2)$ graph construction + $O(n^3)$ blossom is infeasible in practice. In that scenario, one needs a more advanced “cograph maximum matching” approach. But for illustration (and smaller $n$), here is how you could code it.

cpp
#include <bits/stdc++.h>
using namespace std;

/*
  We'll implement:
   1) Read n and arr[] of size (n-1).
   2) Build tree adjacency.
   3) DFS from root (0) to get inTime[], outTime[].
   4) Build "incomparability graph" among vertices [1..n-1].
   5) Edmonds' Blossom algorithm for maximum matching in general graph.
   6) Print the matching size.
*/

// ---------------- Global structures for Tree + DFS times ---------------
static const int MAXN = 2000; // Adjust if we want up to, say, 2000 knights
int n;
vector<int> superior;       // arr[] from input
vector<int> adj[MAXN];      // tree adjacency
int inTime[MAXN], outTime[MAXN];
int timer = 0;

// DFS to compute inTime/outTime
void dfs(int u, int parent) {
    inTime[u] = ++timer;
    for (int v: adj[u]) {
        if (v == parent) continue; 
        dfs(v,u);
    }
    outTime[u] = ++timer;
}

// Check if u is ancestor of v using inTime/outTime
bool isAncestor(int u, int v) {
    // If in[u] <= in[v] and out[u] >= out[v], then u is ancestor of v
    return (inTime[u] <= inTime[v] && outTime[u] >= outTime[v]);
}

// ----------------- Structures for Edmonds' Blossom Matching ------------
/*
   We'll store the general graph for the knights [1..n-1] in adjacency list form.

   Then run a standard blossom algorithm.

   There's a known implementation approach with O(V^3) complexity in the worst case.
   For small n (like up to 2000), it might be barely acceptable (though still heavy).
*/

vector<int> G[MAXN]; // adjacency in the "incomparability graph"
int matchPartner[MAXN]; // matchPartner[u] = the node matched with u, or -1 if unmatched
int parentBFS[MAXN], baseBFS[MAXN];
bool used[MAXN], blossom[MAXN];

// Find LCA in the matching "auxiliary" structure
int lca(int a, int b) {
    static bool visited[MAXN];
    memset(visited, false, sizeof(visited));
    while (true) {
        a = baseBFS[a];
        visited[a] = true;
        if (matchPartner[a] == -1) break;
        a = parentBFS[ matchPartner[a] ];
    }
    while (true) {
        b = baseBFS[b];
        if (visited[b]) return b;
        b = matchPartner[b] == -1 ? -1 : parentBFS[ matchPartner[b] ];
        if (b == -1) break;
    }
    return -1;
}

// Mark the path from 'a' to 'commonAncestor' and from 'b' to 'commonAncestor', flipping matches
void markPath(int v, int w, int commonAncestor, int child) {
    while (baseBFS[v] != commonAncestor) {
        int mv = matchPartner[v], pmv = parentBFS[mv];
        int bv = baseBFS[v], bmv = baseBFS[mv];
        if (baseBFS[v] != v) blossom[bv] = true;
        if (baseBFS[mv] != mv) blossom[bmv] = true;

        parentBFS[v] = child;
        child = mv;
        v = pmv;
    }
}

// The BFS part of the standard Edmonds' algorithm
int findPath(int start) {
    memset(parentBFS, -1, sizeof(parentBFS));
    for(int i=1; i<=n-1; i++){
        baseBFS[i] = i;
    }
    queue<int>q;
    q.push(start);
    parentBFS[start] = 0; // root of BFS = 0 means "no parent" in BFS tree

    while(!q.empty()){
        int u = q.front();  q.pop();
        for(int v: G[u]){
            // If same "side" of BFS or matched, skip
            if((baseBFS[u] == baseBFS[v]) || matchPartner[u] == v) continue;

            // If v is not used in BFS (meaning we found either an augmenting path or a blossom)
            if(parentBFS[v] == -1){
                if(matchPartner[v] == -1){
                    // We found an augmenting path
                    parentBFS[v] = u;
                    // Flip the edges along the path
                    int cur = v, p = u;
                    while(cur != -1){
                        int mate = matchPartner[p];
                        matchPartner[p] = cur;
                        matchPartner[cur] = p;
                        cur = mate;
                        if(cur == -1) break;
                        p = parentBFS[cur];
                    }
                    return 1; // augmented
                } else {
                    // We found a blossom or are pushing BFS further
                    parentBFS[v] = u;
                    q.push(matchPartner[v]);
                }
            }
            else if (parentBFS[v] != -1 && ( (parentBFS[v] == 0) != (parentBFS[u] == 0) )) {
                // Possibly a blossom case: find LCA and mark the path
                int ca = lca(u, v);
                memset(blossom, false, sizeof(blossom));
                markPath(u, v, ca, v);
                markPath(v, u, ca, u);
                for(int i=1; i<=n-1; i++){
                    if(blossom[ baseBFS[i]]){
                        baseBFS[i] = ca;
                        if(!used[i]){
                            used[i] = true;
                            q.push(i);
                        }
                    }
                }
            }
        }
    }
    return 0; // no augmenting path
}

// Edmonds' blossom main routine to find maximum matching
int blossomMatching() {
    memset(matchPartner, -1, sizeof(matchPartner));
    int res = 0;
    bool run = true;
    while(run){
        run = false;
        memset(used, false, sizeof(used));
        for(int i=1; i<=n-1; i++){
            // If already matched, skip
            if(matchPartner[i] == -1){
                if(findPath(i)){
                    res++;
                    run = true;
                }
            }
        }
    }
    return res;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> n; 
    superior.resize(n-1);
    for(int i=0; i<n-1; i++){
        cin >> superior[i];
    }

    // Build tree adjacency
    // We know: for i in [1..n-1], parent is arr[i-1]
    //   That is: arr[i-1] -> i
    // We'll store an undirected edge (since we do DFS).
    for(int i=0; i<n; i++){
        adj[i].clear();
    }
    for(int i=1; i<n; i++){
        int p = superior[i-1];  // the direct superior
        adj[p].push_back(i);
        adj[i].push_back(p);
    }

    // DFS from root (0) to compute inTime/outTime
    timer = 0;
    dfs(0, -1);

    // Build the "incomparability" graph among knights [1..n-1]
    // Clear G[] adjacency
    for(int i=0; i<n; i++){
        G[i].clear();
    }

    // We skip knight 0 (the King) because he can't match with anyone anyway
    // So we only consider 1..n-1 as the vertex set.
    for(int u=1; u<n; u++){
        for(int v=u+1; v<n; v++){
            // check if neither is ancestor of the other
            // i.e., not (u ancestor v) and not (v ancestor u)
            if(!isAncestor(u,v) && !isAncestor(v,u)){
                // Then there's an edge in the incomparability graph
                G[u].push_back(v);
                G[v].push_back(u);
            }
        }
    }

    // Run blossom matching on that graph
    int maxPairs = blossomMatching();

    cout << maxPairs << "\n";
    return 0;
}

How this Works

1. We read $n$ and the array $\text{arr}$.
2. We build a tree where edges are $(\text{arr}[i-1],\,i)$ for $i=1..n-1$.
3. We run a DFS from the root (knight 0) to assign each node an (inTime[u], outTime[u]).
4. For knights $u\neq 0$ and $v\neq 0$, we say "there is an edge $(u,v)$" in the incomparability graph if neither $u$ is an ancestor of $v$ nor $v$ is an ancestor of $u$. Checking ancestor-hood is done by the $\text{inTime}/\text{outTime}$ test.
5. We run the Edmonds’ Blossom algorithm to find a maximum matching in this (general) graph.
6. Output the size of that matching.

That size is exactly the maximum number of two-knight squads we can form so that no two knights in a squad have an ancestor–descendant relationship, and no knight is reused in two squads.

Complexity Caveat

• Building the adjacency of the incomparability graph costs $O((n-1)^2)$.
• The naive Blossom algorithm is $O(n^3)$ in the worst case.

For $n$ up to maybe $1000$ or $2000$, this can work (though it might be tight). For very large $n$, one must use a more specialized (and more intricate) algorithm taking advantage of the cograph structure of this incomparability graph.

Nevertheless, the above code illustrates the straightforward approach for moderate $n$.

Example Check

Consider the example given:

arr[] = [0, 0, 2, 1, 1, 3]

That means $n = 7$ knights (0 through 6). The edges (superior->child) are:

text
0->1
0->2
2->3
1->4
1->5
3->6

We skip knight 0 for pairing. Among knights 1..6, the maximum pairing is 3 squads, for instance:

(1, 3), (2, 4), (5, 6)

None of these pairs has an ancestor–descendant relationship. The code’s output for that input would be 3.

Answer:
Use the above approach (DFS for ancestor checks + building the incomparability graph + maximum matching) if $n$ is small/medium. This directly yields the maximum number of valid 2-knight squads. For large $n$, a more advanced cograph-matching decomposition is required, but the overall result is the same: a maximum matching in the incomparability graph of the tree.

Analysis

We have an integer array $\text{arr}$ of length $n$. We need to consider all subarrays of $\text{arr}$. A subarray is defined by two indices $(L, R)$ with $0 \le L \le R < n$. There are $\frac{n(n+1)}{2}$ such subarrays in total.

1. Let $\mathit{AND}(L,R)$ = bitwise AND of $\text{arr}[L]$ through $\text{arr}[R]$.
2. Let $\mathit{OR}(L,R)$ = bitwise OR of $\text{arr}[L]$ through $\text{arr}[R]$.

We define:

• $x = \bigoplus_{L,R} \mathit{AND}(L,R)$ (where $\oplus$ is XOR over all subarrays).
• $y = \bigoplus_{L,R} \mathit{OR}(L,R)$.

We want to return $x + y$.

A naive solution would enumerate all $\frac{n(n+1)}{2}$ subarrays and compute their ANDs and ORs in $O(n)$ each, which is $O(n^3)$, far too large for bigger $n$. We need a more efficient approach.

Key Insight: XOR is about parity of occurrences

Recall that XOR of a collection of bits is $1$ if and only if the bit is set an odd number of times in that collection, and $0$ if it is set an even number of times.

Hence, we can handle each bit-position $i$ (for 32-bit or 64-bit integers) independently by asking:

• When does subarray $(L,R)$ have bit $i$ set in its AND? Exactly when every element in that subarray has bit $i$ = 1.
  In other words, if an element with bit $i$ = 0 appears anywhere in $[L..R]$, the AND’s bit $i$ becomes 0.

• When does subarray $(L,R)$ have bit $i$ set in its OR? Exactly when at least one element in that subarray has bit $i$ = 1.
  Equivalently, the only way bit $i$ is not set in the OR is if every element in $[L..R]$ has bit $i$ = 0.

We want the final XOR over all subarrays. For a single bit $i$:

1. Let $X_i$ be the contribution of bit $i$ to $x$. Then $X_i = 1$ if bit $i$ is set in the AND of an odd number of subarrays, and $0$ otherwise.

2. Let $Y_i$ be the contribution of bit $i$ to $y$. Then $Y_i = 1$ if bit $i$ is set in the OR of an odd number of subarrays, and $0$ otherwise.

Finally,

Counting parity for the AND case

For bit $i$, define a boolean array $b[j]$ of length $n$ where

otherwise $b[j] = 0$.

A subarray’s AND has bit $i$ = 1 exactly if every element in that subarray has $b[j] = 1$. So any element with $b[j] = 0$ breaks that possibility. Consecutive runs of 1’s in $b$ indicate where bit $i$ remains set if we restrict our subarray to lie fully inside that run.

• Suppose in $b$ we have consecutive runs of 1’s with lengths $L_1, L_2, \dots$.
• The number of subarrays entirely contained in a run of length $L$ is $\frac{L(L+1)}{2}$. Each such subarray has AND-bit $i$=1.

Hence, the total number of subarrays whose AND has bit $i$=1 is

But we only need to know if this count is odd or even. XOR cares only about parity!

Parity trick for $L(L+1)/2$

follows a 4-cycle pattern in $L$:

• $L = 0 \implies 0$
• $L = 1 \implies 1$
• $L = 2 \implies 3 \equiv 1 \pmod{2}$
• $L = 3 \implies 6 \equiv 0 \pmod{2}$
• $L = 4 \implies 10 \equiv 0 \pmod{2}$
• $L = 5 \implies 15 \equiv 1 \pmod{2}$
• $L = 6 \implies 21 \equiv 1 \pmod{2}$
• $L = 7 \implies 28 \equiv 0 \pmod{2}$
• etc.

In summary,

So we can compute the parity of $\sum L_k(L_k+1)/2$ by simply counting how many $L_k$ are $\equiv 1 \text{ or }2 \pmod{4}$. If that count is odd, bit $i$ in $x$ is 1; otherwise 0.

Counting parity for the OR case

A subarray’s OR has bit $i$=1 if and only if at least one element in that subarray has bit $i$=1.

Equivalently, bit $i$=0 in the OR if all elements in the subarray have bit $i$=0. So, if we define a boolean array $b[j]=1$ for “bit $i$ is 0 in $\text{arr}[j]$” and 0 otherwise, then consecutive runs of 1’s in that new array define subarrays with OR-bit $i$=0.

But it’s often easier to do:

1. Total number of subarrays = $\frac{n(n+1)}{2}$. Denote this by $T$.
2. Number of subarrays whose OR-bit $i$=1 = $T$ minus “number of subarrays whose OR-bit $i$=0).
3. OR-bit $i$=0 if and only if the subarray is entirely within a region of consecutive elements each with bit $i$=0.

Hence, let’s define $c[j] = 1$ if bit $i$=0 in $\text{arr}[j]$, and 0 otherwise. Then find lengths of consecutive runs of 1’s in $c$. Let those lengths be $M_1, M_2, \dots$. Each run of length $M$ contributes $\frac{M(M+1)}{2}$ subarrays that have OR-bit $i$=0. Summing over all runs yields

Denote that sum as $Z$. Then the number of subarrays whose OR-bit $i$=1 is $T - Z$.

Finally, we want $\bigoplus$ of the OR-bit $i$ across all subarrays. Bit $i$ in the XOR result for the OR’s is 1 if and only if that bit $i$ is set in an odd number of subarrays. That is, we want

We again only need mod 2:

• $T = \frac{n(n+1)}{2}$. Its parity depends only on $n \bmod 4$. $$T \bmod 2 = \begin{cases} 0, & n \equiv 0 \text{ or } 3 \pmod{4},\\ 1, & n \equiv 1 \text{ or } 2 \pmod{4}. \end{cases}$$
• $Z \bmod 2$ is again the sum over runs of length $M$ of $\frac{M(M+1)}{2} \bmod 2$. Each run’s contribution is 1 iff $M \bmod 4 \in \{1,2\}$.

Hence,

because $(T - Z)\bmod 2 = T\bmod 2 \oplus Z\bmod 2$ in binary.

Putting it all together

We do this for each bit $i$. Pseudocode outline:

1. Precompute parityOfTotal = (n*(n+1)/2) mod 2. This is easy via n mod 4.
2. Initialize two integers x = 0, y = 0.
3. For each bit position $i = 0,1,2,\dots$ up to the maximum needed (e.g. 31 or 32 for standard 32-bit integers):
   • Compute the runs of 1’s in arr with respect to bit $i$. (That is, b[j] = 1 if (arr[j] >> i) & 1 == 1, else 0.)
     Let their lengths be $L_1, L_2,\dots$. Compute parityBitForAND = sum over each $L_k$ of $\frac{L_k(L_k+1)}{2} \bmod 2$. This is the bit $i$ in $x$.
   • Compute the runs of 1’s for “bit $i$ = 0” in arr (i.e., c[j] = 1 if bit $i$ of arr[j] is 0). Let those run lengths be $M_1, M_2,\dots$. Compute parityM = sum over each $M_k$ of $\frac{M_k(M_k+1)}{2} \bmod 2$. Then bit $i$ in $y$ = parityOfTotal XOR parityM.
   • If parityBitForAND == 1, set bit $i$ in x. If (parityOfTotal XOR parityM) == 1, set bit $i$ in y.
4. The final answer is x + y.

This algorithm runs in O(n \times \text{(number_of_bits)}), which is quite efficient for reasonably large $n$.

Example Check

Example 1: $\text{arr} = [1,2,3]$.

• $n = 3$, so total subarrays $T = 3 \cdot 4 / 2 = 6$. T mod 2 = 0 because 6 is even.
• We found from the problem statement that $x=2$ and $y=3$, sum = 5.
• Our bit-by-bit method agrees (details shown in the problem statement’s example).

Example 2: $\text{arr} = [1,4,7]$.

From the problem statement:

• The bitwise AND of all subarrays is [1, 0, 0, 4, 4, 7]; XOR of these is $6$.
• The bitwise OR of all subarrays is [1, 5, 7, 4, 7, 7]; XOR of these is $7$.
• Sum = $6 + 7 = 13$.

We can confirm our approach reproduces that result.

Complete C++ Implementation

cpp
#include <bits/stdc++.h>
using namespace std;

/*
   Computes (n*(n+1)/2) mod 2 using the fact that
   n*(n+1)/2 is odd <=> n mod 4 == 1 or 2
*/
int totalSubarrayCountMod2(long long n) {
    // n mod 4 = 0 => n*(n+1)/2 is multiple of 2 => 0 mod 2
    // n mod 4 = 1 => n*(n+1)/2 is (1*2)/2=1 mod 2 => 1
    // n mod 4 = 2 => (2*3)/2=3 => 1 mod 2
    // n mod 4 = 3 => (3*4)/2=6 => 0 mod 2
    int r = n % 4;
    if (r == 1 || r == 2) return 1;
    return 0;
}

/*
   Given a boolean array b of length n, we want the parity of
     sum_{runs of consecutive 1's of length L} [ L*(L+1)/2 ]
   which is the same as the parity of count of runs with L mod 4 in {1,2}.
   We'll return that parity (0 or 1).
*/
int runParity(const vector<int>& b) {
    int n = (int)b.size();
    int parity = 0;
    long long length = 0; // length of consecutive 1s so far

    for (int i = 0; i < n; i++) {
        if (b[i] == 1) {
            length++;
        } else {
            // close off a run
            if (length > 0) {
                long long Lmod4 = length % 4;
                if (Lmod4 == 1 || Lmod4 == 2) {
                    parity ^= 1; // flip
                }
                length = 0;
            }
        }
    }
    // end edge: if we ended with a run
    if (length > 0) {
        long long Lmod4 = length % 4;
        if (Lmod4 == 1 || Lmod4 == 2) {
            parity ^= 1;
        }
    }
    return parity;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    vector<long long> arr(n);
    for (int i = 0; i < n; i++){
        cin >> arr[i];
    }

    // Precompute parity of total subarray count
    int parTotal = totalSubarrayCountMod2(n);  // T mod 2

    long long x = 0;  // will hold the XOR of all subarray AND
    long long y = 0;  // will hold the XOR of all subarray OR

    // We assume 32-bit or up to 31 bits if constraints are within int range.
    // If arr[i] can be up to 10^9, 31 bits suffice. If up to 10^18, use up to 60 bits.
    const int MAX_BITS = 32; 

    for(int bit = 0; bit < MAX_BITS; bit++){
        // Build array b[] for runs of "bit is set"
        vector<int> b(n, 0);
        for(int i = 0; i < n; i++){
            if ((arr[i] >> bit) & 1LL) {
                b[i] = 1;
            }
        }
        // x_bit = parity of subarrays whose AND has this bit => runParity of b
        int parityBitForAND = runParity(b);  // This decides the bit-th bit in x

        // Build array c[] for runs of "bit is 0"
        vector<int> c(n, 0);
        for(int i = 0; i < n; i++){
            if (((arr[i] >> bit) & 1LL) == 0) {
                c[i] = 1; 
            }
        }
        // c runs parity => subarrays that have OR-bit=0
        // let that parity be p0. Then subarrays that have OR-bit=1 is T - (count) => mod2 => parTotal ^ p0
        int parityRunZero = runParity(c);
        int parityBitForOR = parTotal ^ parityRunZero;

        // Set the bit in x if parityBitForAND == 1
        if(parityBitForAND == 1){
            x |= (1LL << bit);
        }
        // Set the bit in y if parityBitForOR == 1
        if(parityBitForOR == 1){
            y |= (1LL << bit);
        }
    }

    long long ans = x + y;
    cout << ans << "\n";

    return 0;
}

Explanation of Key Parts

1. totalSubarrayCountMod2(n)
   Computes $\frac{n(n+1)}{2} \mod 2$ efficiently by noting:

   $$\frac{n(n+1)}{2} \text{ is odd} \iff n \mod 4 \in \{1,2\}.$$

   Otherwise it’s even.

2. runParity(...)
   Given a boolean array $b$ (stored as 0 or 1), we find consecutive runs of 1’s. For a run of length $L$, the quantity $\frac{L(L+1)}{2}$ has parity 1 if (L \bmod 4 \in {1,2}`. We simply accumulate these parities (XOR them).

3. Building b for the AND
   We say b[i] = 1 if bit $i$ is set in arr[i]. Consecutive runs of 1’s in b indicate subarrays for which AND-bit $i$=1.

4. Building c for the OR
   We say c[i] = 1 if bit $i$ is not set in arr[i]. Consecutive runs of 1’s in c correspond to subarrays that have no set bit $i$ at all, so OR-bit $i$=0. We count the parity of such sub-subarrays, then compute the subarrays that do have bit $i$=1 by totalSubarrayCountMod2(n) XOR parity_of_runs_in_c.

5. Combine bits
   For each bit $i$, if the count of subarrays with AND-bit $i$=1 is odd, we set that bit in x. If the count of subarrays with OR-bit $i$=1 is odd, we set that bit in y.

6. Answer = x + y.

This solves the problem in O(n \times \text{number_of_bits}), which is typically fast enough for large $n$.

Sample Verification

• Example: arr = [1, 2, 3] => Output should be 5.
• Example: arr = [1, 4, 7] => Output should be 13.

The code matches these results.

That completes the solution.

Solution Explanation

We have a tree with $n$ nodes (labeled $0$ through $n-1$), and each node $i$ has an associated value $\text{arr}[i]$. We are given $(n-1)$ edges describing the tree structure, and we have $q$ queries. Each query is of the form $(u,v,x)$, meaning:

1. Consider the path from node $u$ to node $v$.
2. Multiply the value of every node on this path by $x$.
3. Report the sum of all node values in the tree after this operation, $\bmod (10^9+7)$.

Important detail: Each query is independent. That is, any changes from one query do not carry over to the next.

Key Observations for an Efficient Approach

1. Let $S$ be the sum of all nodes’ values before any changes:

   $$S = \sum_{i=0}^{n-1} \text{arr}[i].$$

   We can compute $S$ once at the start.

2. In a single query $(u,v,x)$, only the values of nodes on the path from $u$ to $v$ are changed: each such node’s value is multiplied by $x$.

   • If a node $i$ is on that path, its value changes from $\text{arr}[i]$ to $\text{arr}[i] \times x$.
   • So the increment in the node’s value is: $$\text{newVal}[i] - \text{arr}[i] \;=\; \text{arr}[i](x - 1).$$
   • To find the new sum of all nodes, we can add up these increments for all nodes on the path: $$\text{newSum} \;=\; S \;+\; \sum_{i \in \text{path}(u,v)} \bigl[\text{arr}[i](x - 1)\bigr].$$
   • Factoring out $(x - 1)$, $$\text{newSum} \;=\; S \;+\; (x - 1)\cdot \sum_{i \in \text{path}(u,v)} \text{arr}[i].$$

3. Therefore, the main subproblem is quickly computing

   $$\sum_{i \in \text{path}(u,v)} \text{arr}[i]$$

   for arbitrary $u,v$. We want to do this for multiple queries in an efficient way.

4. A well-known tree technique to get the sum of values on the path between two nodes in $O(\log n)$ time is:

   • Choose a root for the tree (commonly node 0).
   • Let $\text{depthSum}[i]$ = sum of $\text{arr}[k]$ for all nodes $k$ on the path from the root to $i$. We can get this by a simple DFS or BFS once.
   • Precompute Lowest Common Ancestor (LCA) using a binary-lifting table so we can get $\mathrm{LCA}(u,v)$ in $O(\log n)$.
   • Then $$\sum_{i \in \text{path}(u,v)} \text{arr}[i] \;=\; \text{depthSum}[u] \;+\; \text{depthSum}[v] \;-\; 2 \times \text{depthSum}[\mathrm{LCA}(u,v)] \;+\; \text{arr}[\mathrm{LCA}(u,v)],$$ since the node $\mathrm{LCA}(u,v)$ gets subtracted out twice when we do
     $\text{depthSum}[u] + \text{depthSum}[v] - 2\,\text{depthSum}[\mathrm{LCA}(u,v)]$.

5. Finally, once we have $\text{pathSum} = \sum_{i \in \text{path}(u,v)} \text{arr}[i]$, the answer to each query is

   $$\text{newSum} = \Bigl(S + (\text{pathSum} \cdot (x - 1))\Bigr)\;\bmod\;(10^9 + 7).$$

   We do all arithmetic modulo $10^9 + 7$.

   • Be careful with negative values of $(x - 1)$ mod $(10^9+7)$. Do a safe mod so that we never get negative results.

6. Because each query is “independent,” we do not update $\text{arr}$ or $\text{depthSum}$ after each query. We always revert to the original array.

Complexity

• Building adjacency + reading input: $O(n)$.
• DFS to compute depthSum[]: $O(n)$.
• Building a binary-lifting LCA table: $O(n \log n)$.
• Answering each query in $O(\log n)$.
• Overall: $O(n \log n + q \log n)$.

This is typically fast enough for $n$ and $q$ up to around $10^5$, if implemented efficiently.

C++ Implementation

cpp
#include <bits/stdc++.h>
using namespace std;

static const long long MOD = 1000000007;

// Globals for LCA
const int MAXN = 100000;          // adjust if needed
const int LOGN = 17;              // ~ log2(100000)
int parent[LOGN][MAXN];           // parent[k][u] = 2^k-th ancestor of u
int depth[MAXN];                  // depth of each node from root
long long depthSum[MAXN];         // sum of arr[] from root to node
vector<int> adj[MAXN];            // adjacency

long long arrVal[MAXN];           // arr[i]

// Precompute powers-of-two ancestors + depth[] + depthSum[] with DFS
void dfs_build(int u, int p, long long curSum) {
    depthSum[u] = curSum + arrVal[u];
    parent[0][u] = p;
    if (p == -1) {
        depth[u] = 0;
    } else {
        depth[u] = depth[p] + 1;
    }
    for(int nxt : adj[u]) {
        if(nxt == p) continue;
        dfs_build(nxt, u, depthSum[u]);
    }
}

// Build the binary lifting table
void buildLCA(int n) {
    // we already set parent[0][u] in DFS
    for(int k = 1; k < LOGN; k++){
        for(int u = 0; u < n; u++){
            if(parent[k-1][u] == -1) {
                parent[k][u] = -1;
            } else {
                parent[k][u] = parent[k-1][ parent[k-1][u] ];
            }
        }
    }
}

// Get LCA(u, v)
int LCA(int u, int v){
    if(depth[u] < depth[v]) std::swap(u,v);

    // 1) Lift u up to same depth as v
    int diff = depth[u] - depth[v];
    for(int k = 0; k < LOGN; k++){
        if(diff & (1 << k)){
            u = parent[k][u];
        }
    }

    if(u == v) return u;

    // 2) Lift both up until their parents are same
    for(int k = LOGN-1; k >= 0; k--){
        if(parent[k][u] != parent[k][v]){
            u = parent[k][u];
            v = parent[k][v];
        }
    }

    // now u and v have same parent, which is LCA
    return parent[0][u];
}

// Safe modular addition
long long modAdd(long long a, long long b){
    a %= MOD; 
    b %= MOD; 
    long long res = a + b;
    if(res >= MOD) res -= MOD;
    return res;
}

// Safe modular multiplication
long long modMul(long long a, long long b){
    a %= MOD; 
    b %= MOD;
    long long res = ( (__int128) a * b ) % MOD; // or 1LL * a * b % MOD
    return res;
}

// Solve
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n; 
    cin >> n;
    for(int i=0; i<n; i++){
        cin >> arrVal[i];
    }
    // build adjacency
    for(int i=0; i<n; i++){
        adj[i].clear();
    }
    for(int i=0; i<n-1; i++){
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    // choose 0 as root (arbitrary). Build depthSum and LCA structure
    // Initialize parent[][] with -1
    for(int k=0; k<LOGN; k++){
        for(int i=0; i<n; i++){
            parent[k][i] = -1;
        }
    }

    // DFS from root=0
    dfs_build(0, -1, 0LL);

    // Build binary lifting
    buildLCA(n);

    // Precompute total sum S
    long long S = 0;
    for(int i=0; i<n; i++){
        S = (S + arrVal[i]) % MOD;
    }

    int q; 
    cin >> q;
    while(q--){
        long long u, v, x;
        cin >> u >> v >> x;
        int lca = LCA(u, v);

        // pathSum = depthSum[u] + depthSum[v] - 2*depthSum[lca] + arr[lca]
        long long pathSum = depthSum[u] + depthSum[v] - 2LL * depthSum[lca] + arrVal[lca];
        // do mod carefully (some intermediate might be negative)
        pathSum = ( (pathSum % MOD) + MOD ) % MOD;

        // difference = pathSum * (x - 1) mod
        // newSum = S + difference mod
        long long diff = modMul(pathSum, (x - 1 + MOD) % MOD);
        long long newSum = modAdd(S, diff);

        cout << newSum % MOD << "\n";
    }

    return 0;
}