#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int maxn = 3e5 + 10, lg = 20;
vector<pair<long long, long long> > v[maxn];
long long n, w[maxn], rmq[lg + 2][maxn], dp[maxn], up[maxn], W[maxn],
    par[lg + 2][maxn], h[maxn], q;
void dfs_down(int i, int j) {
  h[i] = h[j] + 1;
  dp[i] = w[i];
  for (int y = 0; y < v[i].size(); y++) {
    int x = v[i][y].first;
    if (x - j) {
      W[x] = v[i][y].second;
      dfs_down(x, i);
      dp[i] += max(0LL, dp[x] - 2 * W[x]);
    }
  }
}
long long get_dp(int pes) {
  long long s = 0;
  if (dp[pes] - 2 * W[pes] >= 0 && pes != 1) s = -(dp[pes] - 2 * W[pes]);
  return s;
}
void dfs_up(int i, int j) {
  if (i != 1) {
    up[i] = max(0LL, up[j] + dp[j] + get_dp(i) - 2 * W[i]);
  }
  for (int y = 0; y < v[i].size(); y++) {
    int x = v[i][y].first;
    if (x - j) {
      dfs_up(x, i);
    }
  }
}
void get_par(int i, int j) {
  par[0][i] = j;
  rmq[0][i] = -W[i] + dp[j] + get_dp(i);
  for (int y = 1; y <= lg; y++) {
    int p = par[y - 1][i];
    par[y][i] = par[y - 1][p];
    rmq[y][i] = rmq[y - 1][i] + rmq[y - 1][p];
  }
  for (int y = 0; y < v[i].size(); y++) {
    int x = v[i][y].first;
    if (x - j) {
      get_par(x, i);
    }
  }
}
long long query(int i, int j) {
  if (i == j) return up[i] + dp[i];
  if (h[i] > h[j]) swap(i, j);
  int o = h[j] - h[i];
  int I = i;
  int J = j;
  long long Ans = 0;
  for (int y = 0; y <= lg; y++) {
    if (o & (1 << y)) {
      Ans += rmq[y][j];
      j = par[y][j];
    }
  }
  if (j == i) return Ans + dp[J] + up[i];
  Ans += dp[I] + dp[J];
  for (int y = lg; y > -1; y--) {
    if (par[y][i] != par[y][j])
      Ans += rmq[y][j] + rmq[y][i], i = par[y][i], j = par[y][j];
  }
  int p = par[0][j];
  Ans += up[p] + rmq[0][i] + rmq[0][j] + get_dp(i) + get_dp(j) -
         (dp[p] + get_dp(i) + get_dp(j));
  return Ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int y = 1; y <= n; y++) cin >> w[y];
  for (int y = 1; y < n; y++) {
    int i, j, p;
    cin >> i >> j >> p;
    v[i].push_back(pair<long long, long long>(j, p));
    v[j].push_back(pair<long long, long long>(i, p));
  }
  dfs_down(1, 0);
  dfs_up(1, 0);
  get_par(1, 0);
  while (q--) {
    int i, j;
    cin >> i >> j;
    cout << query(i, j) << endl;
  }
}

### Prompt

Please formulate a CPP solution to the following problem:

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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int maxn = 3e5 + 10, lg = 20;
vector<pair<long long, long long> > v[maxn];
long long n, w[maxn], rmq[lg + 2][maxn], dp[maxn], up[maxn], W[maxn],
    par[lg + 2][maxn], h[maxn], q;
void dfs_down(int i, int j) {
  h[i] = h[j] + 1;
  dp[i] = w[i];
  for (int y = 0; y < v[i].size(); y++) {
    int x = v[i][y].first;
    if (x - j) {
      W[x] = v[i][y].second;
      dfs_down(x, i);
      dp[i] += max(0LL, dp[x] - 2 * W[x]);
    }
  }
}
long long get_dp(int pes) {
  long long s = 0;
  if (dp[pes] - 2 * W[pes] >= 0 && pes != 1) s = -(dp[pes] - 2 * W[pes]);
  return s;
}
void dfs_up(int i, int j) {
  if (i != 1) {
    up[i] = max(0LL, up[j] + dp[j] + get_dp(i) - 2 * W[i]);
  }
  for (int y = 0; y < v[i].size(); y++) {
    int x = v[i][y].first;
    if (x - j) {
      dfs_up(x, i);
    }
  }
}
void get_par(int i, int j) {
  par[0][i] = j;
  rmq[0][i] = -W[i] + dp[j] + get_dp(i);
  for (int y = 1; y <= lg; y++) {
    int p = par[y - 1][i];
    par[y][i] = par[y - 1][p];
    rmq[y][i] = rmq[y - 1][i] + rmq[y - 1][p];
  }
  for (int y = 0; y < v[i].size(); y++) {
    int x = v[i][y].first;
    if (x - j) {
      get_par(x, i);
    }
  }
}
long long query(int i, int j) {
  if (i == j) return up[i] + dp[i];
  if (h[i] > h[j]) swap(i, j);
  int o = h[j] - h[i];
  int I = i;
  int J = j;
  long long Ans = 0;
  for (int y = 0; y <= lg; y++) {
    if (o & (1 << y)) {
      Ans += rmq[y][j];
      j = par[y][j];
    }
  }
  if (j == i) return Ans + dp[J] + up[i];
  Ans += dp[I] + dp[J];
  for (int y = lg; y > -1; y--) {
    if (par[y][i] != par[y][j])
      Ans += rmq[y][j] + rmq[y][i], i = par[y][i], j = par[y][j];
  }
  int p = par[0][j];
  Ans += up[p] + rmq[0][i] + rmq[0][j] + get_dp(i) + get_dp(j) -
         (dp[p] + get_dp(i) + get_dp(j));
  return Ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int y = 1; y <= n; y++) cin >> w[y];
  for (int y = 1; y < n; y++) {
    int i, j, p;
    cin >> i >> j >> p;
    v[i].push_back(pair<long long, long long>(j, p));
    v[j].push_back(pair<long long, long long>(i, p));
  }
  dfs_down(1, 0);
  dfs_up(1, 0);
  get_par(1, 0);
  while (q--) {
    int i, j;
    cin >> i >> j;
    cout << query(i, j) << endl;
  }
}

```