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;
long long w[300005];
vector<pair<int, long long> > nei[300005];
long long dp[300005], dp2[20][300005], up[300005];
int fa[20][300005];
int dep[300005];
long long la[300005];
void dfs(int v, int pa, int depth) {
  dep[v] = depth;
  fa[0][v] = pa;
  dp[v] = w[v];
  for (int i = 0; i < nei[v].size(); i++) {
    int u = nei[v][i].first;
    long long co = nei[v][i].second;
    if (u == pa) continue;
    la[u] = co;
    dfs(u, v, depth + 1);
    dp[v] += max(0ll, dp[u] - 2 * co);
  }
}
void dfs2(int v, int pa) {
  for (int i = 0; i < nei[v].size(); i++) {
    int u = nei[v][i].first;
    long long co = nei[v][i].second;
    if (u == pa) continue;
    if (dp[u] - 2 * co > 0) {
      up[u] = max(up[v] + dp[v] - dp[u], 0ll);
      dp2[0][u] = dp[v] - dp[u] + co;
    } else {
      up[u] = max(up[v] + dp[v] - 2 * co, 0ll);
      dp2[0][u] = dp[v] - co;
    }
    dfs2(u, v);
  }
}
pair<long long, int> lca(int x, int y) {
  long long res = 0, de = 0;
  if (dep[x] != dep[y]) {
    if (dep[x] > dep[y]) swap(x, y);
    for (int i = 19; i >= 0; i--) {
      if (dep[fa[i][y]] > dep[x]) res += dp2[i][y], y = fa[i][y];
    }
    res += dp2[0][y];
    y = fa[0][y];
  }
  if (x == y) return make_pair(res, x);
  for (int i = 19; i >= 0; i--) {
    if (fa[i][x] != fa[i][y]) {
      res += dp2[i][x] + dp2[i][y];
      x = fa[i][x];
      y = fa[i][y];
    }
  }
  res += dp2[0][x] + dp2[0][y];
  x = fa[0][x];
  return make_pair(res - dp[x], x);
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", &w[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    long long c;
    scanf("%d %d %I64d", &x, &y, &c);
    x--, y--;
    nei[x].push_back(make_pair(y, c));
    nei[y].push_back(make_pair(x, c));
  }
  dfs(0, 0, 0);
  dp2[0][0] = 0;
  up[0] = 0;
  dfs2(0, 0);
  for (int i = 1; i < 20; i++) {
    for (int j = 0; j < n; j++) {
      dp2[i][j] = dp2[i - 1][j] + dp2[i - 1][fa[i - 1][j]];
      fa[i][j] = fa[i - 1][fa[i - 1][j]];
    }
  }
  while (q--) {
    int x, y;
    scanf("%d %d", &x, &y);
    x--, y--;
    if (dep[x] > dep[y]) swap(x, y);
    pair<long long, int> tmp = lca(x, y);
    if (x == y)
      printf("%I64d\n", dp[x] + up[x]);
    else if (tmp.second == x)
      printf("%I64d\n", dp[y] + tmp.first + up[x]);
    else
      printf("%I64d\n", tmp.first + dp[x] + dp[y] + up[tmp.second]);
  }
}