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 arr[300005], dep[300005];
long long paredge[300005];
long long top_extra[300005];
long long sub_extra[300005];
vector<pair<long long, long long>> adj[300005];
struct Node {
  long long pos;
  long long path_val, extra;
  Node() { pos = 0, path_val = 0, extra = 0; }
};
Node par[300005][20];
void dfs(long long v, long long p) {
  par[v][0].pos = p;
  par[v][0].path_val = arr[v] - paredge[v];
  sub_extra[v] = arr[v];
  dep[v] = dep[p] + 1;
  for (auto u : adj[v]) {
    if (u.first == p) continue;
    paredge[u.first] = u.second;
    dfs(u.first, v);
    sub_extra[v] += max(0ll, sub_extra[u.first] - 2 * u.second);
  }
}
void dfs2(long long v, long long p) {
  top_extra[v] =
      max(0ll, (sub_extra[p] - max(0ll, sub_extra[v] - 2 * paredge[v])) +
                   top_extra[p] - 2 * paredge[v]);
  for (auto u : adj[v]) {
    if (u.first == p) continue;
    dfs2(u.first, v);
  }
}
void answer(long long u, long long v) {
  if (dep[u] < dep[v]) swap(u, v);
  long long extra = 0;
  long long path = 0;
  long long tmpu = u, tmpv = v;
  for (long long i = 20 - 1; i >= 0; i--) {
    if (dep[par[u][i].pos] >= dep[v]) {
      if (tmpu == u) {
        extra += par[u][i].extra;
      } else {
        extra += par[u][i].extra - sub_extra[u] + arr[u];
      }
      path += par[u][i].path_val;
      u = par[u][i].pos;
    }
  }
  if (u == v) {
    cout << extra + path + top_extra[u] + arr[u] << endl;
    return;
  }
  for (long long i = 20 - 1; i >= 0; i--) {
    if (par[u][i].pos != par[v][i].pos) {
      if (tmpu == u) {
        extra += par[u][i].extra;
      } else {
        extra += par[u][i].extra - sub_extra[u] + arr[u];
      }
      path += par[u][i].path_val;
      u = par[u][i].pos;
      if (tmpv == v) {
        extra += par[v][i].extra;
      } else {
        extra += par[v][i].extra - sub_extra[v] + arr[v];
      }
      path += par[v][i].path_val;
      v = par[v][i].pos;
    }
  }
  long long lca = par[u][0].pos;
  path += par[u][0].path_val + par[v][0].path_val;
  if (tmpu == u) {
    extra += sub_extra[u] - arr[u];
  }
  if (tmpv == v) {
    extra += sub_extra[v] - arr[v];
  }
  cout << top_extra[lca] + path + extra + sub_extra[lca] -
              max(0ll, sub_extra[u] - 2 * paredge[u]) -
              max(0ll, sub_extra[v] - 2 * paredge[v])
       << endl;
}
void solve() {
  long long n, q;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  for (long long i = 1; i < n; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    adj[u].push_back({v, w});
    adj[v].push_back({u, w});
  }
  dfs(1, 0);
  dfs2(1, 0);
  for (long long i = 1; i <= n; i++) {
    par[i][0].extra = sub_extra[i] - arr[i] + sub_extra[par[i][0].pos] -
                      arr[par[i][0].pos] -
                      max(0ll, sub_extra[i] - 2 * paredge[i]);
  }
  for (long long j = 1; j < 20; j++) {
    for (long long i = 1; i <= n; i++) {
      par[i][j].pos = par[par[i][j - 1].pos][j - 1].pos;
      par[i][j].path_val =
          par[i][j - 1].path_val + par[par[i][j - 1].pos][j - 1].path_val;
      par[i][j].extra = par[i][j - 1].extra +
                        par[par[i][j - 1].pos][j - 1].extra -
                        sub_extra[par[i][j - 1].pos] + arr[par[i][j - 1].pos];
    }
  }
  while (q--) {
    long long u, v;
    cin >> u >> v;
    if (u == v) {
      cout << sub_extra[v] + top_extra[v] << endl;
      continue;
    }
    answer(u, v);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  while (t--) {
    solve();
  }
}