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 mxN = 3e5, mxQ = 4e5;
int n, q, eu[mxN - 1], ev[mxN - 1], dep[mxN], fo[mxN], dt[20][2 * mxN - 1], dti,
    pew[mxN];
long long a[mxN], w[mxN - 1], smp[mxN], smp2[mxN], pmp[mxN];
vector<int> adj[mxN];
void dfs1(int u, int p) {
  dep[u] = p == -1 ? 0 : dep[p] + 1;
  fo[u] = dti;
  dt[0][dti++] = u;
  for (int e : adj[u]) {
    int v = u ^ eu[e] ^ ev[e];
    if (v == p) continue;
    pew[v] = w[e];
    dfs1(v, u);
    dt[0][dti++] = u;
    smp[u] += max(smp[v], 0ll);
  }
  smp[u] += a[u] - 2 * pew[u];
}
inline int lg2(int x) { return 31 - __builtin_clz(x); }
inline int lca(int u, int v) {
  if (u == v) return u;
  if ((u = fo[u]) > (v = fo[v])) swap(u, v);
  int k = lg2(v - u);
  return dep[dt[k][u]] < dep[dt[k][v - (1 << k) + 1]] ? dt[k][u]
                                                      : dt[k][v - (1 << k) + 1];
}
void dfs2(int u, int p) {
  smp2[u] += smp[u] + pew[u] - max(smp[u], 0ll);
  long long ssmp = pmp[u];
  for (int e : adj[u]) {
    int v = u ^ eu[e] ^ ev[e];
    if (v == p) continue;
    smp2[v] += smp2[u];
    ssmp += max(smp[v], 0ll);
  }
  for (int e : adj[u]) {
    int v = u ^ eu[e] ^ ev[e];
    if (v == p) continue;
    pmp[v] = max(a[u] - 2 * w[e] + ssmp - max(smp[v], 0ll), 0ll);
    dfs2(v, u);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n - 1; ++i) {
    cin >> eu[i] >> ev[i] >> w[i], --eu[i], --ev[i];
    adj[eu[i]].push_back(i);
    adj[ev[i]].push_back(i);
  }
  dfs1(0, -1);
  for (int i = 1; i <= lg2(2 * n - 1); ++i)
    for (int j = 0; j < 2 * n - (1 << i); ++j)
      dt[i][j] = dep[dt[i - 1][j]] < dep[dt[i - 1][j + (1 << (i - 1))]]
                     ? dt[i - 1][j]
                     : dt[i - 1][j + (1 << (i - 1))];
  dfs2(0, -1);
  for (int i = 0; i < q; ++i) {
    int qu, qv;
    cin >> qu >> qv, --qu, --qv;
    int qw = lca(qu, qv);
    long long ans =
        smp2[qu] + smp2[qv] - 2 * smp2[qw] + smp[qw] + 2 * pew[qw] + pmp[qw];
    cout << ans << "\n";
  }
}