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 const M = 3e5 + 10, M2 = 1e6 + 10, mod = 998244353, mod2 = 1e9 + 9,
                inf = 1e9 + 30;
long long a[M], dp[M], ans[M], sum[M][21], par[M][21], bad[M][21], h[M], sv[M],
    up[M];
vector<long long> adj[M], edg[M];
void dfs_down(long long v, long long l) {
  dp[v] = a[v];
  for (long long i = 0; i < adj[v].size(); i++) {
    long long u = adj[v][i];
    if (u == l) continue;
    dfs_down(u, v);
    long long tmp = dp[u] - edg[v][i] * 2;
    if (tmp > 0) dp[v] += tmp;
  }
  sv[v] = dp[v];
}
void dfs_up(long long v, long long l, long long yal) {
  h[v] = h[l] + 1;
  if (l != 0) {
    long long now = dp[l];
    long long tmp = dp[v] - yal * 2;
    if (tmp > 0) now -= tmp;
    long long tmp2 = now - yal * 2;
    if (tmp2 > 0) dp[v] += tmp2, up[v] = tmp2;
  }
  par[v][0] = l;
  sum[v][0] = sv[l] - yal;
  long long tmp = sv[v] - yal * 2;
  if (tmp > 0) bad[v][0] = tmp;
  for (long long j = 1; j <= 20; j++) {
    par[v][j] = par[par[v][j - 1]][j - 1];
    sum[v][j] = sum[par[v][j - 1]][j - 1] + sum[v][j - 1];
    bad[v][j] = bad[par[v][j - 1]][j - 1] + bad[v][j - 1];
  }
  for (long long i = 0; i < adj[v].size(); i++) {
    long long u = adj[v][i];
    if (u == l) continue;
    dfs_up(u, v, edg[v][i]);
  }
}
long long qu(long long v, long long u) {
  bool flag = 0;
  if (v + u == 11) flag = 1;
  long long ans = sv[v] + sv[u];
  if (h[v] < h[u]) swap(v, u);
  for (long long i = 20; i >= 0; i--) {
    if (h[v] - (1 << i) >= h[u]) ans += sum[v][i] - bad[v][i], v = par[v][i];
  }
  if (v == u) {
    ans -= sv[u];
    ans += up[v];
    return ans;
  }
  for (long long i = 20; i >= 0; i--) {
    if (par[v][i] != par[u][i]) {
      ans += sum[v][i] - bad[v][i], ans += sum[u][i] - bad[u][i], v = par[v][i],
                                                                  u = par[u][i];
    }
  }
  ans += sum[v][0] - bad[v][0];
  ans += (sum[u][0] - bad[u][0]);
  v = par[v][0];
  ans -= sv[v];
  ans += up[v];
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, q;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n - 1; i++) {
    long long v, u, w;
    cin >> v >> u >> w;
    edg[v].push_back(w);
    edg[u].push_back(w);
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  dfs_down(1, 0);
  dfs_up(1, 0, 0);
  while (q--) {
    long long v, u;
    cin >> v >> u;
    cout << qu(v, u) << "\n";
  }
}