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;
const int maxn = 3e5 + 5;
const int logn = 20;
int n, q;
int a[maxn];
vector<pair<int, int> > adj[maxn];
int par[maxn][logn], lev[maxn];
long long f[maxn], g[maxn], h[maxn], s[maxn], ss[maxn];
void dfs(int u, int p = 0) {
  ss[u] += a[u];
  par[u][0] = p, lev[u] = lev[p] + 1;
  for (int i = 1; i < logn; ++i) {
    par[u][i] = par[par[u][i - 1]][i - 1];
  }
  f[u] = a[u];
  for (int i = 0; i < adj[u].size(); ++i) {
    int v = adj[u][i].first;
    int w = adj[u][i].second;
    if (v != p) {
      ss[v] = ss[u] - w;
      dfs(v, u);
      f[u] += max(0ll, f[v] - 2 * w);
    }
  }
}
int lca(int u, int v) {
  if (lev[u] < lev[v]) {
    swap(u, v);
  }
  for (int i = logn - 1; i >= 0; --i) {
    if (lev[u] - (1 << i) >= lev[v]) {
      u = par[u][i];
    }
  }
  if (u == v) {
    return u;
  }
  for (int i = logn - 1; i >= 0; --i) {
    if (par[u][i] != par[v][i]) {
      u = par[u][i], v = par[v][i];
    }
  }
  return par[u][0];
}
void calc(int u, int p = 0) {
  for (int i = 0; i < adj[u].size(); ++i) {
    int v = adj[u][i].first;
    int w = adj[u][i].second;
    g[u] += max(0ll, f[v] - 2 * w);
    if (v != p) {
      s[u] += max(0ll, f[v] - 2 * w);
    }
  }
  for (int i = 0; i < adj[u].size(); ++i) {
    int v = adj[u][i].first;
    int w = adj[u][i].second;
    if (v != p) {
      h[v] = max(0ll, f[v] - 2 * w);
      s[v] = s[u] - h[v];
      f[u] = g[u] - h[v] + a[u];
      calc(v, u);
    }
  }
}
long long query(int u, int v) {
  int w = lca(u, v);
  return ss[u] + ss[v] - 2 * ss[w] + a[w] + s[u] + s[v] - 2 * s[w] + g[w];
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    adj[u].push_back(pair<int, int>(v, w));
    adj[v].push_back(pair<int, int>(u, w));
  }
  dfs(1);
  calc(1);
  while (q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    printf("%I64d\n", query(u, v));
  }
  return 0;
}