#include <bits/stdc++.h>
using namespace std;
int n, q, x, y, z;
int baba[19][300005], der[300005], a[300005], dadc[300005];
long long res[19][300005], dp[300005], updp[300005];
vector<pair<int, int> > v[300005];
long long getres(int x, int lca) {
  long long ret = 0;
  int dx = der[x];
  for (int i = 19 - 1; i >= 0; i--) {
    if (dx - (1 << (i)) + 1 >= der[lca]) {
      dx -= (1 << (i));
      if (dx >= der[lca])
        ret -=
            dadc[baba[i][x]] + max(0ll, dp[baba[i][x]] - 2 * dadc[baba[i][x]]);
      ret += res[i][x];
      x = baba[1][baba[i][x]];
    }
  }
  return ret;
}
int getlca(int x, int y) {
  if (der[x] < der[y]) swap(x, y);
  int dx = der[x];
  int dy = der[y];
  for (int i = 19 - 1; i >= 0; i--) {
    if (dx - (1 << (i)) >= dy) {
      dx -= (1 << (i));
      x = baba[1][baba[i][x]];
    }
  }
  if (x == y) return x;
  for (int i = 19 - 1; i >= 0; i--) {
    if (baba[1][baba[i][x]] != baba[1][baba[i][y]]) {
      x = baba[1][baba[i][x]];
      y = baba[1][baba[i][y]];
    }
  }
  return baba[1][x];
}
void buildlca() {
  for (int i = 1; i < 19; i++) {
    if (i > 1) {
      for (int j = 1; j <= n; j++) {
        baba[i][j] = baba[1][baba[i - 1][baba[i - 1][j]]];
      }
    }
    for (int j = 1; j <= n; j++) {
      res[i][j] = res[i - 1][j] + res[i - 1][baba[1][baba[i - 1][j]]] -
                  max(0ll, dp[baba[i - 1][j]] - 2 * dadc[baba[i - 1][j]]) -
                  dadc[baba[i - 1][j]];
    }
  }
}
void dfs2(int node, int ata) {
  for (pair<int, int> e : v[node]) {
    int go = e.first;
    int cost = e.second;
    if (go == ata) continue;
    updp[go] = max(
        0ll, dp[node] - max(0ll, dp[go] - 2 * cost) + updp[node] - 2 * cost);
    dfs2(go, node);
  }
}
void dfs1(int node, int ata) {
  dp[node] = a[node];
  baba[0][node] = node;
  baba[1][node] = ata;
  der[node] = der[ata] + 1;
  for (pair<int, int> e : v[node]) {
    int go = e.first;
    int cost = e.second;
    if (go == ata) continue;
    dadc[go] = cost;
    dfs1(go, node);
    dp[node] += max(0ll, -2 * cost + dp[go]);
  }
  res[0][node] = dp[node];
}
long long solve(int x, int y) {
  int lca = getlca(x, y);
  long long res1 = getres(x, lca);
  long long res2 = getres(y, lca);
  return res1 + res2 - dp[lca] + max(0ll, updp[lca]);
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i < n; i++) {
    scanf("%d %d %d", &x, &y, &z);
    v[x].push_back({y, z});
    v[y].push_back({x, z});
  }
  dfs1(1, 0);
  dfs2(1, 0);
  buildlca();
  while (q--) {
    scanf("%d %d", &x, &y);
    printf("%lld\n", solve(x, y));
  }
}

### 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>
using namespace std;
int n, q, x, y, z;
int baba[19][300005], der[300005], a[300005], dadc[300005];
long long res[19][300005], dp[300005], updp[300005];
vector<pair<int, int> > v[300005];
long long getres(int x, int lca) {
  long long ret = 0;
  int dx = der[x];
  for (int i = 19 - 1; i >= 0; i--) {
    if (dx - (1 << (i)) + 1 >= der[lca]) {
      dx -= (1 << (i));
      if (dx >= der[lca])
        ret -=
            dadc[baba[i][x]] + max(0ll, dp[baba[i][x]] - 2 * dadc[baba[i][x]]);
      ret += res[i][x];
      x = baba[1][baba[i][x]];
    }
  }
  return ret;
}
int getlca(int x, int y) {
  if (der[x] < der[y]) swap(x, y);
  int dx = der[x];
  int dy = der[y];
  for (int i = 19 - 1; i >= 0; i--) {
    if (dx - (1 << (i)) >= dy) {
      dx -= (1 << (i));
      x = baba[1][baba[i][x]];
    }
  }
  if (x == y) return x;
  for (int i = 19 - 1; i >= 0; i--) {
    if (baba[1][baba[i][x]] != baba[1][baba[i][y]]) {
      x = baba[1][baba[i][x]];
      y = baba[1][baba[i][y]];
    }
  }
  return baba[1][x];
}
void buildlca() {
  for (int i = 1; i < 19; i++) {
    if (i > 1) {
      for (int j = 1; j <= n; j++) {
        baba[i][j] = baba[1][baba[i - 1][baba[i - 1][j]]];
      }
    }
    for (int j = 1; j <= n; j++) {
      res[i][j] = res[i - 1][j] + res[i - 1][baba[1][baba[i - 1][j]]] -
                  max(0ll, dp[baba[i - 1][j]] - 2 * dadc[baba[i - 1][j]]) -
                  dadc[baba[i - 1][j]];
    }
  }
}
void dfs2(int node, int ata) {
  for (pair<int, int> e : v[node]) {
    int go = e.first;
    int cost = e.second;
    if (go == ata) continue;
    updp[go] = max(
        0ll, dp[node] - max(0ll, dp[go] - 2 * cost) + updp[node] - 2 * cost);
    dfs2(go, node);
  }
}
void dfs1(int node, int ata) {
  dp[node] = a[node];
  baba[0][node] = node;
  baba[1][node] = ata;
  der[node] = der[ata] + 1;
  for (pair<int, int> e : v[node]) {
    int go = e.first;
    int cost = e.second;
    if (go == ata) continue;
    dadc[go] = cost;
    dfs1(go, node);
    dp[node] += max(0ll, -2 * cost + dp[go]);
  }
  res[0][node] = dp[node];
}
long long solve(int x, int y) {
  int lca = getlca(x, y);
  long long res1 = getres(x, lca);
  long long res2 = getres(y, lca);
  return res1 + res2 - dp[lca] + max(0ll, updp[lca]);
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i < n; i++) {
    scanf("%d %d %d", &x, &y, &z);
    v[x].push_back({y, z});
    v[y].push_back({x, z});
  }
  dfs1(1, 0);
  dfs2(1, 0);
  buildlca();
  while (q--) {
    scanf("%d %d", &x, &y);
    printf("%lld\n", solve(x, y));
  }
}

```