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 MN = 400005, inf = 1000000005, MOD = 1000000007, LG = 20;
const long long INF = 1000000000000000005LL;
int hop[LG][MN];
int used[MN], pre[MN], post[MN];
long long nad[MN], pod[MN], C[MN], up[MN], val[MN];
long long suma[LG][MN];
vector<pair<int, int> > G[MN];
int czas;
void dfs(int x) {
  used[x] = 1;
  pre[x] = ++czas;
  pod[x] = val[x];
  for (auto v : G[x])
    if (!used[v.first]) {
      hop[0][v.first] = x;
      dfs(v.first);
      up[v.first] = max(0LL, pod[v.first] - 2LL * v.second);
      C[v.first] = v.second;
      pod[x] += up[v.first];
    }
  post[x] = ++czas;
}
void dfs2(int x) {
  used[x] = 2;
  nad[x] += val[x];
  for (auto v : G[x])
    if (used[v.first] != 2) {
      nad[v.first] =
          max(0LL, nad[x] - 2LL * v.second + pod[x] - up[v.first] - val[x]);
      dfs2(v.first);
    }
}
bool child(int a, int b) { return (pre[a] >= pre[b] && post[a] <= post[b]); }
long long skacz(int a, int b) {
  if (child(b, a)) return 0LL;
  int s = LG - 1;
  long long ret = pod[a];
  while (s >= 0) {
    if (child(b, hop[s][a])) {
      --s;
    } else {
      ret -= pod[a];
      ret += suma[s][a];
      a = hop[s][a];
    }
  }
  ret -= C[a];
  ret -= up[a];
  return ret;
}
int lca(int a, int b) {
  if (child(b, a)) return a;
  if (child(a, b)) return b;
  int s = LG - 1;
  while (s >= 0) {
    if (child(b, hop[s][a]))
      --s;
    else
      a = hop[s][a];
  }
  return hop[0][a];
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%lld", &val[i]);
  for (int i = 1; i < n; ++i) {
    int a, b;
    long long w;
    scanf("%d%d%lld", &a, &b, &w);
    G[a].push_back({b, w});
    G[b].push_back({a, w});
  }
  dfs(1);
  dfs2(1);
  hop[0][1] = 1;
  for (int j = 1; j <= n; ++j)
    suma[0][j] = pod[j] + pod[hop[0][j]] - up[j] - C[j];
  for (int i = 1; i < LG; ++i)
    for (int j = 1; j <= n; ++j) {
      hop[i][j] = hop[i - 1][hop[i - 1][j]];
      suma[i][j] =
          suma[i - 1][j] + suma[i - 1][hop[i - 1][j]] - pod[hop[i - 1][j]];
    }
  while (q--) {
    int a, b;
    scanf("%d%d", &a, &b);
    long long res = 0LL;
    res += skacz(a, b);
    res += skacz(b, a);
    int l = lca(a, b);
    res += (pod[l] + nad[l] - val[l]);
    printf("%lld\n", res);
  }
  return 0;
}