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 N = 3e5 + 10;
int test;
int n;
int v[N][2];
long long wv[N], we[N];
vector<int> e[N];
int depth[N], pv[N][20];
int pe[N], q[N];
void BFS() {
  int top = 1, bot = 1;
  q[1] = 1;
  while (top <= bot) {
    int xa = q[top++];
    for (auto i : e[xa])
      if (pe[xa] != i) {
        int xb = v[i][0] + v[i][1] - xa;
        q[++bot] = xb;
        pe[xb] = i;
      }
  }
}
void prepare() {
  cin >> n >> test;
  for (int i = 1; i <= n; i++) cin >> wv[i];
  for (int i = 1; i < n; i++) {
    int xa, xb;
    cin >> xa >> xb >> we[i];
    e[xa].push_back(i);
    e[xb].push_back(i);
    v[i][0] = xa;
    v[i][1] = xb;
  }
  BFS();
  for (int i = 2; i <= n; i++) {
    int xa = q[i];
    int par_xa = v[pe[xa]][0] + v[pe[xa]][1] - xa;
    depth[xa] = depth[par_xa] + 1;
    pv[xa][0] = par_xa;
  }
}
long long f1[N], fp[N];
long long pathf[N][20];
void solve() {
  for (int i = n; i >= 1; i--) {
    int xa = q[i];
    for (auto edge : e[xa])
      if (edge != pe[xa]) {
        int xb = v[edge][0] + v[edge][1] - xa;
        f1[xa] += max(0LL, f1[xb] - 2 * we[edge] + wv[xb]);
      }
  }
  for (int i = 1; i <= n; i++) {
    int xa = q[i];
    long long sum = fp[xa] + f1[xa];
    for (auto edge : e[xa])
      if (edge != pe[xa]) {
        int xb = v[edge][0] + v[edge][1] - xa;
        fp[xb] = max(0LL, sum - max(0LL, f1[xb] - 2 * we[edge] + wv[xb]) -
                              2 * we[edge] + wv[xa]);
      }
    for (auto edge : e[xa])
      if (edge != pe[xa]) {
        int xb = v[edge][0] + v[edge][1] - xa;
        pathf[xb][0] = wv[xb] + f1[xa] - we[edge] -
                       max(0LL, f1[xb] - 2 * we[edge] + wv[xb]);
      }
  }
  for (int j = 1; j < 20; j++)
    for (int i = 1; i <= n; i++) {
      pv[i][j] = pv[pv[i][j - 1]][j - 1];
      pathf[i][j] = pathf[i][j - 1] + pathf[pv[i][j - 1]][j - 1];
    }
}
long long query(int x, int y) {
  if (depth[x] < depth[y]) swap(x, y);
  long long ans = f1[x];
  for (int i = 0, val = depth[x] - depth[y]; val; i++, val >>= 1)
    if (val & 1) {
      ans += pathf[x][i];
      x = pv[x][i];
    }
  if (x == y) {
    ans += wv[y] + fp[y];
    return ans;
  }
  ans += f1[y];
  for (int i = int(log2(depth[x])); i >= 0; i--)
    if (pv[x][i] != pv[y][i]) {
      ans += pathf[x][i] + pathf[y][i];
      x = pv[x][i], y = pv[y][i];
    }
  int z = pv[x][0];
  ans += wv[x] + wv[y] + wv[z];
  ans -= we[pe[x]] + we[pe[y]];
  ans += f1[z] + fp[z] - max(0LL, f1[x] - 2 * we[pe[x]] + wv[x]) -
         max(0LL, f1[y] - 2 * we[pe[y]] + wv[y]);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  prepare();
  solve();
  while (test--) {
    int xa, xb;
    cin >> xa >> xb;
    cout << query(xa, xb) << '\n';
  }
}