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 = 300005;
struct edge {
  int to, next, v;
} e[N * 2];
int n, Q, v[N];
int head[N], tot;
int fa[N][20], dep[N];
long long f[N], F[N], g[N], G[N];
long long S1[N], S2[N];
void add(int x, int y, int v) {
  e[++tot] = (edge){y, head[x], v};
  head[x] = tot;
}
void dfs(int x) {
  for (int i = head[x]; i; i = e[i].next)
    if (e[i].to != fa[x][0]) {
      fa[e[i].to][0] = x;
      dep[e[i].to] = dep[x] + 1;
      dfs(e[i].to);
    }
}
void init1(int x) {
  for (int i = head[x]; i; i = e[i].next)
    if (e[i].to != fa[x][0]) {
      init1(e[i].to);
      F[e[i].to] = max(0ll, f[e[i].to] + v[e[i].to] - 2 * e[i].v);
      f[x] += F[e[i].to];
    }
}
void init2(int x) {
  for (int i = head[x]; i; i = e[i].next)
    if (e[i].to != fa[x][0]) {
      g[e[i].to] = f[x] + G[x] - F[e[i].to];
      G[e[i].to] = max(0ll, g[e[i].to] + v[x] - 2 * e[i].v);
      init2(e[i].to);
    }
}
void init3(int x) {
  S1[x] += f[x] + G[x];
  for (int i = head[x]; i; i = e[i].next)
    if (e[i].to != fa[x][0]) {
      S1[e[i].to] = S1[x];
      S2[e[i].to] = S2[x] - F[e[i].to] - G[e[i].to] - e[i].v + v[e[i].to];
      init3(e[i].to);
    }
}
int LCA(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  int tmp = dep[x] - dep[y];
  for (int i = (int)(0); i <= (int)(18); i++)
    if (tmp & (1 << i)) x = fa[x][i];
  for (int i = (int)(18); i >= (int)(0); i--)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return x == y ? x : fa[x][0];
}
int main() {
  scanf("%d%d", &n, &Q);
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &v[i]);
  for (int i = (int)(1); i <= (int)(n - 1); i++) {
    int x, y, v;
    scanf("%d%d%d", &x, &y, &v);
    add(x, y, v);
    add(y, x, v);
  }
  dfs(1);
  init1(1);
  init2(1);
  init3(1);
  for (int i = (int)(1); i <= (int)(18); i++)
    for (int j = (int)(1); j <= (int)(n); j++)
      fa[j][i] = fa[fa[j][i - 1]][i - 1];
  while (Q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    int L = LCA(x, y);
    long long ans = S1[x] + S1[y] - S1[L] - S1[fa[L][0]];
    ans += S2[x] + S2[y] - 2 * S2[L];
    ans += v[L];
    printf("%lld\n", ans);
  }
}