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;
#pragma warning(disable : 4996)
struct Edge {
  int to, w;
};
const int N = 3e5 + 5, L = 20;
vector<Edge> v[N];
int a[N], tin[N], tout[N], p[N][L];
long long maxPath[N], maxUpPath[N], sum[N], pt[N], ancw[N];
int timer = 0;
bool upper(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int getBeforeLca(int a, int b) {
  if (upper(a, b)) return -1;
  for (int i = L - 1; i >= 0; i--) {
    if (!upper(p[a][i], b)) a = p[a][i];
  }
  return a;
}
void dfs(int node, int anc) {
  p[node][0] = anc;
  for (int i = 1; i < L; i++) {
    p[node][i] = p[p[node][i - 1]][i - 1];
  }
  tin[node] = timer++;
  sum[node] = maxPath[node] = a[node];
  for (auto e : v[node]) {
    if (e.to != anc) {
      ancw[e.to] = e.w;
      dfs(e.to, node);
      maxPath[e.to] = max(0ll, maxPath[e.to] - e.w * 2);
      maxPath[node] += maxPath[e.to];
      sum[node] += maxPath[e.to];
    }
  }
  tout[node] = timer;
}
void deep(int node, long long x, long long y, int anc) {
  maxUpPath[node] = x - y;
  x += sum[node];
  for (auto e : v[node]) {
    if (e.to != anc) {
      pt[e.to] = pt[node] - e.w + sum[node] - maxPath[e.to];
      long long nx = x - e.w * 2 - maxPath[e.to];
      deep(e.to, nx, min(y, nx), node);
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b, w;
    cin >> a >> b >> w;
    a--, b--;
    v[a].push_back({b, w});
    v[b].push_back({a, w});
  }
  dfs(0, 0);
  deep(0, 0, 0, -1);
  for (int i = 0; i < q; i++) {
    int x, y, xx, yy, lca;
    cin >> x >> y;
    x--, y--;
    xx = getBeforeLca(x, y);
    yy = getBeforeLca(y, x);
    lca = xx == -1 ? x : p[xx][0];
    long long ans = maxUpPath[lca] + sum[lca];
    if (x != lca) {
      ans += sum[x] - maxPath[xx] + pt[x] - pt[xx] - ancw[xx];
    }
    if (y != lca) {
      ans += sum[y] - maxPath[yy] + pt[y] - pt[yy] - ancw[yy];
    }
    cout << ans << "\n";
  }
  return 0;
}