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 N = 3e5 + 5;
int n, q;
vector<pair<int, int> > nei[N];
long long ds[N], u[N], us[N], dg[N];
int a[N], d[N], P[N];
int p[25][N];
long long uv[25][N];
void dfs(int x) {
  for (int i = 0; i < nei[x].size(); i++) {
    int to = nei[x][i].first;
    if (to != p[0][x]) {
      p[0][to] = x;
      d[to] = d[x] + 1;
      P[to] = nei[x][i].second;
      dfs(to);
      ds[x] += max(0ll, ds[to] + a[to] - 2 * P[to]);
    }
  }
  dg[x] = max(0ll, ds[x] + a[x] - 2 * P[x]);
  for (int i = 0; i < nei[x].size(); i++) {
    int to = nei[x][i].first;
    if (to != p[0][x]) u[to] = ds[x] - max(0ll, dg[to]);
  }
}
void dfs2(int x) {
  if (x == 1)
    us[1] = a[1];
  else
    us[x] = max(0ll, ds[p[0][x]] - dg[x] + us[p[0][x]] - P[x] * 2) + a[x];
  for (int i = 0; i < nei[x].size(); i++)
    if (nei[x][i].first != p[0][x]) dfs2(nei[x][i].first);
}
int lca(int x, int y) {
  if (d[x] > d[y]) swap(x, y);
  for (int i = 24; i >= 0; i--)
    if ((d[y] - d[x]) & (1 << i)) y = p[i][y];
  if (x == y) return x;
  for (int i = 24; i >= 0; i--)
    if (p[i][x] != p[i][y]) x = p[i][x], y = p[i][y];
  return p[0][x];
}
long long calc(int &x, int d) {
  long long ret = 0;
  for (int i = 0; i < 24; i++)
    if (d & (1 << i)) ret += uv[i][x], x = p[i][x];
  return ret;
}
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++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    nei[x].push_back(make_pair(y, z));
    nei[y].push_back(make_pair(x, z));
  }
  p[0][1] = 1;
  dfs(1);
  dfs2(1);
  for (int i = 1; i <= n; i++) uv[0][i] = u[i] + a[i] - P[i];
  for (int i = 0; i < 24; i++)
    for (int j = 1; j <= n; j++) p[i + 1][j] = p[i][p[i][j]];
  for (int i = 0; i < 24; i++)
    for (int j = 1; j <= n; j++) uv[i + 1][j] = uv[i][j] + uv[i][p[i][j]];
  while (q--) {
    int x, y, l;
    scanf("%d%d", &x, &y);
    l = lca(x, y);
    long long ans = us[l] + ds[l];
    if (x != l) ans += ds[x] + calc(x, d[x] - d[l] - 1) + a[x] - P[x] - dg[x];
    if (y != l) ans += ds[y] + calc(y, d[y] - d[l] - 1) + a[y] - P[y] - dg[y];
    printf("%lld\n", ans);
  }
  return 0;
}