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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MAXN = 4e5 + 100;
int N, Q, a[MAXN];
vector<pair<int, int> > g[MAXN];
const int MAXK = 22;
int depth[MAXN], pa[MAXK][MAXN];
long long answer[MAXK][MAXN];
long long dp[MAXN];
void dfs(int v = 0, int p = -1) {
  pa[0][v] = p;
  dp[v] = a[v];
  for (auto [u, w] : g[v]) {
    if (u ^ p) {
      depth[u] = depth[v] + 1;
      dfs(u, v);
      dp[v] = max(dp[v], dp[v] + dp[u] - 2 * w);
    }
  }
  for (auto [u, w] : g[v]) {
    if (u ^ p) {
      answer[0][u] = dp[v] - max(0ll, dp[u] - 2 * w);
    }
  }
}
long long dp2[MAXN];
void dfs2(int v = 0, int p = -1) {
  for (auto [u, w] : g[v]) {
    if (u ^ p) {
      dp2[u] = max(dp2[u], dp2[v] + answer[0][u] - 2 * w);
      dfs2(u, v);
    }
  }
  for (auto [u, w] : g[v]) {
    if (u ^ p) {
      answer[0][u] -= w;
    }
  }
}
void solve() {
  cin >> N >> Q;
  for (int i = (0); i < (N); i++) cin >> a[i];
  for (int i = (0); i < (N - 1); i++) {
    int u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    g[u].push_back({v, w});
    g[v].push_back({u, w});
  }
  dfs();
  dfs2();
  for (int k = (0); k < (MAXK - 1); k++) {
    for (int i = (0); i < (N); i++) {
      if (pa[k][i] == -1) {
        pa[k + 1][i] = pa[k][i];
        answer[k + 1][i] = answer[k][i];
      } else {
        pa[k + 1][i] = pa[k][pa[k][i]];
        answer[k + 1][i] = answer[k][i] + answer[k][pa[k][i]];
      }
    }
  }
  for (int q = (0); q < (Q); q++) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    if (u == v) {
      cout << dp[v] + dp2[v] << "\n";
      continue;
    }
    long long ans = 0;
    if (depth[u] < depth[v]) swap(u, v);
    int x = u, diff = depth[u] - depth[v];
    for (int k = (MAXK - 1); k >= (0); k--) {
      if (diff >> k & 1) {
        ans += answer[k][x];
        x = pa[k][x];
      }
    }
    if (x == v) {
      ans += dp[u] + dp2[v];
      cout << ans << "\n";
      continue;
    }
    ans += dp[u];
    ans += dp[v];
    u = x;
    for (int k = (MAXK - 1); k >= (0); k--) {
      if (pa[k][u] != pa[k][v]) {
        ans += answer[k][u];
        ans += answer[k][v];
        u = pa[k][u];
        v = pa[k][v];
      }
    }
    ans += answer[0][u];
    ans += answer[0][v];
    ans -= dp[pa[0][u]];
    ans += dp2[pa[0][u]];
    cout << ans << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  solve();
  return 0;
}