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 N, Q;
map<long long, long long> adj[300000 + 1];
long long depth[300000 + 1];
long long dp[300000 + 1];
long long dp2[300000 + 1];
long long up[300000 + 1][20 + 1];
long long edgeSum[300000 + 1][20 + 1];
long long dpSum[300000 + 1][20 + 1];
long long f(long long u, long long v) {
  return max(dp2[v] - 2 * adj[u][v], 0LL);
}
void dfs1(long long node, long long par) {
  for (auto i : adj[node])
    if (i.first != par) {
      dfs1(i.first, node);
      dp2[node] += f(node, i.first);
    }
}
void dfs2(long long node, long long par) {
  dp[node] = dp2[node];
  for (auto i : adj[node]) {
    if (i.first == par) continue;
    dp2[node] -= f(node, i.first);
    dp2[i.first] += f(i.first, node);
    dfs2(i.first, node);
    dp2[i.first] -= f(i.first, node);
    dp2[node] += f(node, i.first);
  }
}
void dfs3(long long node, long long par) {
  depth[node] = depth[up[node][0] = par] + 1;
  dpSum[node][0] = dp2[node] - f(par, node);
  for (long long i = 1; i <= 20; i++) {
    dpSum[node][i] = dpSum[node][i - 1] + dpSum[up[node][i - 1]][i - 1];
    edgeSum[node][i] = edgeSum[node][i - 1] + edgeSum[up[node][i - 1]][i - 1];
    up[node][i] = up[up[node][i - 1]][i - 1];
  }
  for (auto i : adj[node])
    if (i.first != par) {
      edgeSum[i.first][0] = i.second;
      dfs3(i.first, node);
    }
}
pair<long long, long long> jmp(long long x, long long k) {
  long long ret = 0;
  for (long long i = 0; i <= 20; i++)
    if ((k >> i) & 1) {
      ret += dpSum[x][i];
      ret -= edgeSum[x][i];
      x = up[x][i];
    }
  return {x ? x : -1, ret};
}
long long query(long long a, long long b) {
  if (depth[a] < depth[b]) swap(a, b);
  long long ret = 0;
  pair<long long, long long> ans = jmp(a, depth[a] - depth[b]);
  a = ans.first, ret = ans.second;
  assert(a != -1);
  if (a == b) return ret + dp[a];
  for (long long i = 20; i >= 0; i--) {
    long long newA = up[a][i], newB = up[b][i];
    if (newA != newB) {
      ret += dpSum[a][i] + dpSum[b][i];
      ret -= edgeSum[a][i] + edgeSum[b][i];
      a = newA, b = newB;
    }
  }
  return ret + dpSum[a][0] + dpSum[b][0] - edgeSum[a][0] - edgeSum[b][0] +
         dp[up[a][0]];
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> N >> Q;
  for (long long i = 1; i <= N; i++) cin >> dp2[i];
  for (long long i = 1; i < N; i++) {
    long long u, v, c;
    cin >> u >> v >> c;
    adj[u][v] = c;
    adj[v][u] = c;
  }
  dfs1(1, 0);
  dfs2(1, 0);
  dfs3(1, 0);
  for (long long i = 1; i <= Q; i++) {
    long long a, b;
    cin >> a >> b;
    cout << query(a, b) << "\n";
  }
  return 0;
}