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;
int n, m;
int a[300005];
vector<pair<int, int> > nei[300005];
vector<int> son[300005];
int ceng[300005];
int fa[20][300005];
long long facos[300005], fac[20][300005];
long long dp[300005], dp2[300005];
long long ds[300005];
void dfs(int now, int lst) {
  ceng[now] = ceng[lst] + 1;
  fa[0][now] = lst;
  for (int i = 1; i < 20; i++) {
    fa[i][now] = fa[i - 1][fa[i - 1][now]];
  }
  for (int i = 0; i < nei[now].size(); i++) {
    int to = nei[now][i].first;
    if (to == lst) {
      continue;
    }
    son[now].push_back(to);
    dfs(to, now);
    facos[to] = nei[now][i].second;
  }
}
void dfsdown(int now) {
  ds[now] = 0;
  for (int i = 0; i < son[now].size(); i++) {
    int to = son[now][i];
    dfsdown(to);
    ds[now] += dp[to];
  }
  dp[now] = max(0ll, ds[now] + a[now] - 2 * facos[now]);
}
void dfsup(int now) {
  dp2[now] =
      (!fa[0][now] ? 0ll
                   : max(0ll, dp2[fa[0][now]] + ds[fa[0][now]] - dp[now] +
                                  a[fa[0][now]] - 2 * facos[now]));
  for (int i = 0; i < son[now].size(); i++) {
    dfsup(son[now][i]);
  }
}
int lca(int a, int b) {
  if (ceng[a] > ceng[b]) {
    swap(a, b);
  }
  int toup = ceng[b] - ceng[a];
  for (int i = 0; i < 20; i++) {
    if ((toup >> i) & 1) {
      b = fa[i][b];
    }
  }
  if (a == b) {
    return a;
  }
  for (int i = 19; ~i; i--) {
    if (fa[i][a] != fa[i][b]) {
      a = fa[i][a];
      b = fa[i][b];
    }
  }
  return fa[0][a];
}
long long calc(int &a, int ceng) {
  long long ans = 0;
  for (int i = 19; ~i; i--) {
    if ((ceng >> i) & 1) {
      ans += fac[i][a];
      a = fa[i][a];
    }
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 1; i < n; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    nei[a].push_back(make_pair(b, c));
    nei[b].push_back(make_pair(a, c));
  }
  dfs(1, 0);
  dfsdown(1);
  dfsup(1);
  for (int i = 1; i <= n; i++) {
    fac[0][i] = (!fa[0][i] ? 0 : ds[fa[0][i]] - dp[i] + a[fa[0][i]] - facos[i]);
  }
  for (int i = 1; i < 20; i++) {
    for (int j = 1; j <= n; j++) {
      fac[i][j] = fac[i - 1][j] + fac[i - 1][fa[i - 1][j]];
    }
  }
  while (m--) {
    int u, v;
    scanf("%d%d", &u, &v);
    int y = lca(u, v);
    long long ans = 0;
    if (u != y) {
      ans += a[u] + ds[u] + calc(u, ceng[u] - ceng[y] - 1);
    }
    if (v != y) {
      ans += a[v] + ds[v] + calc(v, ceng[v] - ceng[y] - 1);
    }
    ans += (a[y] + dp2[y] + ds[y] - (u == y ? 0 : dp[u] + facos[u]) -
            (v == y ? 0 : dp[v] + facos[v]));
    printf("%lld\n", ans);
  }
  return 0;
}