#include <bits/stdc++.h>
using namespace std;
vector<vector<pair<int, int> > > adj(300001);
int a[300001];
long long dp[300001], dpToRootDetour[300001];
int start[300001], finish[300001], t_time[300001], indx[300001],
    ind = 0, timer = 0, T[19 + 1][2 * 300001], lg[2 * 300001];
void BuildLCASparseTable(int n) {
  n = (n << 1);
  for (int i = 2; i < n; i++) lg[i] = 1 + lg[i / 2];
  for (int k = 1; (1 << k) < n; k++)
    for (int i = 1, j = 1 + (1 << (k - 1)), m = n - (1 << k) + 1; i < m;
         i++, j++)
      T[k][i] = min(T[k - 1][i], T[k - 1][j]);
}
int LCA(int x, int y) {
  x = indx[x], y = indx[y];
  if (x > y) swap(x, y);
  int k = lg[y - x + 1];
  k = min(T[k][x], T[k][y - (1 << k) + 1]);
  int z = t_time[k];
  return z;
}
void DFS1(int i, int p) {
  start[i] = ++timer;
  t_time[timer] = i;
  indx[i] = ++ind;
  T[0][ind] = start[i];
  for (auto e : adj[i]) {
    int k, w;
    tie(k, w) = e;
    if (k == p) {
      continue;
    }
    DFS1(k, i);
    dp[i] += max(0ll, dp[k] - 2 * w);
    T[0][++ind] = start[i];
  }
  dp[i] += a[i];
  finish[i] = timer;
}
void DFS2(int i, int p) {
  for (auto e : adj[i]) {
    int k, w;
    tie(k, w) = e;
    if (k == p) {
      continue;
    }
    dpToRootDetour[k] = dpToRootDetour[i] - max(0ll, dp[k] - 2 * w) - w + dp[k];
    dp[k] = max(0ll, dp[i] - max(0ll, dp[k] - 2 * w) - 2 * w) + dp[k];
    DFS2(k, i);
  }
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  for (int x = 1; x < n; ++x) {
    int i, j, w;
    scanf("%d%d%d", &i, &j, &w);
    adj[i].push_back(make_pair(j, w));
    adj[j].push_back(make_pair(i, w));
  }
  DFS1(1, 0);
  DFS2(1, 0);
  BuildLCASparseTable(n);
  while (q--) {
    int i, j;
    scanf("%d%d", &i, &j);
    int k = LCA(i, j);
    long long ans =
        dpToRootDetour[i] + dpToRootDetour[j] - 2 * dpToRootDetour[k] + dp[k];
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

You are given a weighted tree (undirected connected graph with no cycles, loops or multiple edges) with n vertices. The edge \\{u_j, v_j\} has weight w_j. Also each vertex i has its own value a_i assigned to it.

Let's call a path starting in vertex u and ending in vertex v, where each edge can appear no more than twice (regardless of direction), a 2-path. Vertices can appear in the 2-path multiple times (even start and end vertices).

For some 2-path p profit Pr(p) = ∑_{v ∈ distinct vertices in  p}{a_v} - ∑_{e ∈ distinct edges in  p}{k_e ⋅ w_e}, where k_e is the number of times edge e appears in p. That is, vertices are counted once, but edges are counted the number of times they appear in p.

You are about to answer m queries. Each query is a pair of vertices (qu, qv). For each query find 2-path p from qu to qv with maximal profit Pr(p).

Input

The first line contains two integers n and q (2 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ q ≤ 4 ⋅ 10^5) — the number of vertices in the tree and the number of queries.

The second line contains n space-separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the values of the vertices.

Next n - 1 lines contain descriptions of edges: each line contains three space separated integers u_i, v_i and w_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i, 1 ≤ w_i ≤ 10^9) — there is edge \\{u_i, v_i\} with weight w_i in the tree.

Next q lines contain queries (one per line). Each query contains two integers qu_i and qv_i (1 ≤ qu_i, qv_i ≤ n) — endpoints of the 2-path you need to find.

Output

For each query print one integer per line — maximal profit Pr(p) of the some 2-path p with the corresponding endpoints.

Example

Input

7 6
6 5 5 3 2 1 2
1 2 2
2 3 2
2 4 1
4 5 1
6 4 2
7 3 25
1 1
4 4
5 6
6 4
3 4
3 7


Output

9
9
9
8
12
-14

Note

Explanation of queries: 

  1. (1, 1) — one of the optimal 2-paths is the following: 1 → 2 → 4 → 5 → 4 → 2 → 3 → 2 → 1. Pr(p) = (a_1 + a_2 + a_3 + a_4 + a_5) - (2 ⋅ w(1,2) + 2 ⋅ w(2,3) + 2 ⋅ w(2,4) + 2 ⋅ w(4,5)) = 21 - 2 ⋅ 12 = 9. 
  2. (4, 4): 4 → 2 → 1 → 2 → 3 → 2 → 4. Pr(p) = (a_1 + a_2 + a_3 + a_4) - 2 ⋅ (w(1,2) + w(2,3) + w(2,4)) = 19 - 2 ⋅ 10 = 9. 
  3. (5, 6): 5 → 4 → 2 → 3 → 2 → 1 → 2 → 4 → 6. 
  4. (6, 4): 6 → 4 → 2 → 1 → 2 → 3 → 2 → 4. 
  5. (3, 4): 3 → 2 → 1 → 2 → 4. 
  6. (3, 7): 3 → 2 → 1 → 2 → 4 → 5 → 4 → 2 → 3 → 7. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<vector<pair<int, int> > > adj(300001);
int a[300001];
long long dp[300001], dpToRootDetour[300001];
int start[300001], finish[300001], t_time[300001], indx[300001],
    ind = 0, timer = 0, T[19 + 1][2 * 300001], lg[2 * 300001];
void BuildLCASparseTable(int n) {
  n = (n << 1);
  for (int i = 2; i < n; i++) lg[i] = 1 + lg[i / 2];
  for (int k = 1; (1 << k) < n; k++)
    for (int i = 1, j = 1 + (1 << (k - 1)), m = n - (1 << k) + 1; i < m;
         i++, j++)
      T[k][i] = min(T[k - 1][i], T[k - 1][j]);
}
int LCA(int x, int y) {
  x = indx[x], y = indx[y];
  if (x > y) swap(x, y);
  int k = lg[y - x + 1];
  k = min(T[k][x], T[k][y - (1 << k) + 1]);
  int z = t_time[k];
  return z;
}
void DFS1(int i, int p) {
  start[i] = ++timer;
  t_time[timer] = i;
  indx[i] = ++ind;
  T[0][ind] = start[i];
  for (auto e : adj[i]) {
    int k, w;
    tie(k, w) = e;
    if (k == p) {
      continue;
    }
    DFS1(k, i);
    dp[i] += max(0ll, dp[k] - 2 * w);
    T[0][++ind] = start[i];
  }
  dp[i] += a[i];
  finish[i] = timer;
}
void DFS2(int i, int p) {
  for (auto e : adj[i]) {
    int k, w;
    tie(k, w) = e;
    if (k == p) {
      continue;
    }
    dpToRootDetour[k] = dpToRootDetour[i] - max(0ll, dp[k] - 2 * w) - w + dp[k];
    dp[k] = max(0ll, dp[i] - max(0ll, dp[k] - 2 * w) - 2 * w) + dp[k];
    DFS2(k, i);
  }
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  for (int x = 1; x < n; ++x) {
    int i, j, w;
    scanf("%d%d%d", &i, &j, &w);
    adj[i].push_back(make_pair(j, w));
    adj[j].push_back(make_pair(i, w));
  }
  DFS1(1, 0);
  DFS2(1, 0);
  BuildLCASparseTable(n);
  while (q--) {
    int i, j;
    scanf("%d%d", &i, &j);
    int k = LCA(i, j);
    long long ans =
        dpToRootDetour[i] + dpToRootDetour[j] - 2 * dpToRootDetour[k] + dp[k];
    printf("%lld\n", ans);
  }
  return 0;
}

```