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>
template <typename... Args>
void pr(Args... args) {}
using namespace std;
using ll = long long;
const int MM = 4e5 + 5, LOG = 20;
int n, q;
ll in[MM];
vector<pair<int, int>> adj[MM];
int dep[MM];
int par[MM];
ll pw[MM];
ll dp[MM], dpall[MM], tmp[MM];
int sp[LOG][MM];
ll val[MM];
ll path[MM];
int getlca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = LOG - 1; i >= 0; i--) {
    if (~sp[i][u] and dep[sp[i][u]] >= dep[v]) u = sp[i][u];
  }
  if (u == v) return u;
  for (int i = LOG - 1; i >= 0; i--) {
    if (~sp[i][u] and ~sp[i][v] and sp[i][u] != sp[i][v]) {
      u = sp[i][u];
      v = sp[i][v];
    }
  }
  return sp[0][u];
}
tuple<int, int, int> getnodes(int a, int b) {
  int lca = getlca(a, b);
  for (int i = LOG - 1; i >= 0; i--) {
    if (dep[a] > dep[lca] + (1 << i)) {
      a = sp[i][a];
    }
  }
  for (int i = LOG - 1; i >= 0; i--) {
    if (dep[b] > dep[lca] + (1 << i)) {
      b = sp[i][b];
    }
  }
  return {lca, a, b};
}
void dfs(int cur) {
  dp[cur] = in[cur];
  for (auto [u, w] : adj[cur]) {
    if (u == par[cur]) continue;
    dep[u] = dep[cur] + 1;
    par[u] = cur;
    pw[u] = w;
    dfs(u);
    dp[cur] += max(0LL, dp[u] - 2 * w);
  }
  tmp[cur] = dp[cur];
}
void go(int cur) {
  sp[0][cur] = par[cur];
  val[cur] = val[par[cur]] + dp[cur] - max(0LL, dp[cur] - 2 * pw[cur]);
  dpall[cur] = tmp[cur];
  path[cur] = path[par[cur]] - pw[cur];
  for (auto [u, w] : adj[cur]) {
    if (u == par[cur]) continue;
    tmp[cur] -= max(0LL, tmp[u] - 2 * w);
    tmp[u] += max(0LL, tmp[cur] - 2 * w);
    go(u);
    tmp[u] -= max(0LL, tmp[cur] - 2 * w);
    tmp[cur] += max(0LL, tmp[u] - 2 * w);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> in[i];
  for (int i = 0, a, b, c; i < n - 1; i++) {
    cin >> a >> b >> c;
    adj[a].emplace_back(b, c);
    adj[b].emplace_back(a, c);
  }
  memset(sp, -1, sizeof sp);
  dep[1] = 1;
  dfs(1);
  go(1);
  for (int i = 1; i < LOG; i++) {
    for (int j = 1; j <= n; j++) {
      int u = sp[i - 1][j];
      if (~u) {
        sp[i][j] = sp[i - 1][u];
      }
    }
  }
  while (q--) {
    int a, b;
    cin >> a >> b;
    auto [c, d, e] = getnodes(a, b);
    ll ans = val[a] + val[b] - 2 * val[c] + dpall[c];
    ll p = path[a] + path[b] - 2 * path[c];
    ans += p;
    cout << ans << '\n';
  }
}