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>
const long long N = 300300;
const long long Q = 2e18;
const long long mod = 998244353;
const long long MAGIC = 30;
using namespace std;
int n;
int tim;
int dip[N];
int tin[N];
int tout[N];
long long s[N];
long long A[N];
long long B[N];
long long a[N];
long long e[N];
long long d[N];
long long u[N];
long long f[N];
long long er[N];
int up[N][20];
vector<int> v[N];
void dfs(int x, int p) {
  tin[x] = ++tim;
  up[x][0] = p;
  for (int i = 1; i < 20; i++) {
    up[x][i] = up[up[x][i - 1]][i - 1];
  }
  d[x] = a[x];
  for (int h : v[x]) {
    int y = A[h] ^ B[h] ^ x;
    if (y == p) {
      continue;
    }
    dip[y] = dip[x] + 1;
    s[y] = s[x] - e[h];
    dfs(y, x);
    d[x] += max(0ll, d[y] - e[h] - e[h]);
  }
  for (int h : v[x]) {
    int y = A[h] ^ B[h] ^ x;
    if (y == p) {
      continue;
    }
    f[y] = d[x] - max(0ll, d[y] - e[h] - e[h]);
    er[y] = max(0ll, d[y] - e[h] - e[h]);
  }
  tout[x] = ++tim;
}
void go(int x, int p) {
  for (int h : v[x]) {
    int y = A[h] ^ B[h] ^ x;
    if (y == p) {
      continue;
    }
    u[y] = max(0ll, u[x] + f[y] - e[h] - e[h]);
    f[y] += f[x];
    go(y, x);
  }
}
bool is(int x, int y) { return tin[x] <= tin[y] && tout[x] >= tout[y]; }
int lca(int x, int y) {
  if (is(x, y)) {
    return x;
  }
  if (is(y, x)) {
    return y;
  }
  for (int i = 19; i >= 0; i--) {
    if (!is(up[x][i], y)) {
      x = up[x][i];
    }
  }
  return up[x][0];
}
int go_up(int x, int g) {
  for (int i = 19; i >= 0; i--) {
    if (g >= (1 << i)) {
      g -= (1 << i);
      x = up[x][i];
    }
  }
  return x;
}
long long get(int x, int y) {
  long long profit = 0;
  int p = lca(x, y);
  profit += d[p];
  if (x != p) {
    int g = go_up(x, dip[x] - dip[p] - 1);
    profit += f[x] - f[g];
    profit += d[x] - er[g];
  }
  if (y != p) {
    int g = go_up(y, dip[y] - dip[p] - 1);
    profit += f[y] - f[g];
    profit += d[y] - er[g];
  }
  profit += s[x] + s[y] - s[p] - s[p];
  profit += u[p];
  return profit;
}
void solve() {
  int q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i < n; i++) {
    cin >> A[i] >> B[i] >> e[i];
    v[A[i]].push_back(i);
    v[B[i]].push_back(i);
  }
  dfs(1, 1);
  go(1, 1);
  for (int i = 1; i <= q; i++) {
    int x, y;
    cin >> x >> y;
    cout << get(x, y) << "\n";
  }
}
bool mtest = false;
int main() {
  ios_base::sync_with_stdio(0);
  int TE = 1;
  if (mtest) cin >> TE;
  while (TE--) solve();
  return 0;
}