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. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 300010;
int i, j, k;
int n, q, x, y, z, xx, yy;
int f1[maxn], f2[maxn * 2], f4[maxn * 2], th;
int a[maxn];
long long dp[maxn], f[maxn * 2], f3[maxn * 2], ans[maxn];
int g[21][maxn], gg[maxn], dept[maxn];
long long tot[maxn], anss;
int read() {
  int tot = 0, fh = 1;
  char c = getchar();
  while ((c < '0') || (c > '9')) {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while ((c >= '0') && (c <= '9')) {
    tot = tot * 10 + c - '0';
    c = getchar();
  }
  return tot * fh;
}
void add(int x, int y, int z) {
  th++;
  f2[th] = y;
  f3[th] = z;
  f4[th] = f1[x];
  f1[x] = th;
}
void dfs(int x, int fa) {
  int i = f1[x], j;
  dept[x] = dept[fa] + 1;
  g[0][x] = fa;
  for (j = 1; j <= 20; j++) g[j][x] = g[j - 1][g[j - 1][x]];
  dp[x] = a[x];
  while (i != 0) {
    if (f2[i] != fa) {
      dfs(f2[i], x);
      f[i] = dp[f2[i]];
      if (i % 2 == 1)
        gg[f2[i]] = i + 1;
      else
        gg[f2[i]] = i - 1;
      if (dp[x] < dp[x] - f3[i] * 2 + dp[f2[i]])
        tot[f2[i]] = tot[f2[i]] + f3[i];
      else
        tot[f2[i]] = -f3[i] + dp[f2[i]];
      dp[x] = max(dp[x], dp[x] - f3[i] * 2 + dp[f2[i]]);
    }
    i = f4[i];
  }
}
long long dfs2(int x, int fa) {
  int t;
  if (fa == 0) {
    return max(dp[x], dp[x] + dfs2(g[0][x], gg[x]) - 2 * f3[gg[x]]);
  }
  if (f[fa] != 0)
    return f[fa];
  else {
    if (fa % 2 == 1)
      t = f2[fa + 1];
    else
      t = f2[fa - 1];
    if (dp[t] - 2 * f3[fa] > 0)
      f[fa] = dp[x] - (dp[t] - 2 * f3[fa]);
    else
      f[fa] = dp[x];
    if (g[0][x] != 0) {
      f[fa] = max(f[fa], f[fa] + dfs2(g[0][x], gg[x]) - 2 * f3[gg[x]]);
    }
    return f[fa];
  }
}
void dfs1(int x, int fa) {
  int i = f1[x];
  tot[x] = tot[x] + tot[fa];
  while (i != 0) {
    if (f2[i] != fa) {
      dfs1(f2[i], x);
    }
    i = f4[i];
  }
}
int lca(int x, int y) {
  int i;
  if (dept[x] < dept[y]) swap(x, y);
  for (i = 20; i >= 0; i--) {
    if (dept[g[i][x]] >= dept[y]) {
      x = g[i][x];
    }
  }
  if (x == y) return x;
  for (i = 20; i >= 0; i--) {
    if (g[i][x] != g[i][y]) {
      x = g[i][x];
      y = g[i][y];
    }
  }
  return g[0][x];
}
int getkth(int x, int k) {
  int i = 1, j = 0;
  while (k != 0) {
    if (k % (i * 2) / i == 1) {
      k = k - i;
      x = g[j][x];
    }
    i = i * 2;
    j++;
  }
  return x;
}
int main() {
  n = read();
  q = read();
  for (i = 1; i <= n; i++) {
    a[i] = read();
  }
  for (i = 1; i < n; i++) {
    x = read();
    y = read();
    z = read();
    add(x, y, z);
    add(y, x, z);
  }
  dfs(1, 0);
  ans[1] = dp[1];
  for (i = 2; i <= n; i++) {
    ans[i] = dfs2(i, 0);
  }
  dfs1(1, 0);
  for (i = 1; i <= q; i++) {
    x = read();
    y = read();
    if (x == y) {
      printf("%I64d\n", ans[x]);
      continue;
    }
    z = lca(x, y);
    if (z == y) swap(x, y);
    if (x == z) {
      z = getkth(y, dept[y] - dept[x] - 1);
      anss = tot[y] - tot[z];
      anss = anss + dp[z];
      anss = anss - f3[gg[z]];
      if (-f3[gg[z]] * 2 + dp[z] >= 0)
        anss = anss + ans[x] - (-f3[gg[z]] * 2 + dp[z]);
      else
        anss = anss + ans[x];
      printf("%I64d\n", anss);
    } else {
      xx = getkth(x, dept[x] - dept[z] - 1);
      yy = getkth(y, dept[y] - dept[z] - 1);
      anss = 0;
      anss = anss + tot[x] - tot[xx];
      anss = anss + tot[y] - tot[yy];
      anss = anss + dp[xx];
      anss = anss + dp[yy];
      anss = anss - f3[gg[xx]] - f3[gg[yy]];
      anss = anss + ans[z];
      if (-f3[gg[xx]] * 2 + dp[xx] >= 0)
        anss = anss - (-f3[gg[xx]] * 2 + dp[xx]);
      if (-f3[gg[yy]] * 2 + dp[yy] >= 0)
        anss = anss - (-f3[gg[yy]] * 2 + dp[yy]);
      printf("%I64d\n", anss);
    }
  }
  return 0;
}