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;
struct edge {
  long long to, w, next;
} e[600010];
long long n, cnt, Q, a[300010], fa[300010][30], head[300010], deep[300010],
    f[300010], g[300010], w[300010][30];
inline void add(long long u, long long v, long long w) {
  cnt++;
  e[cnt].to = v;
  e[cnt].w = w;
  e[cnt].next = head[u];
  head[u] = cnt;
}
inline void dp(long long u, long long from) {
  f[u] = a[u];
  for (long long i = head[u]; i; i = e[i].next) {
    long long v = e[i].to;
    if (v == from) continue;
    dp(v, u);
    f[u] += max(0ll, f[v] - (e[i].w << 1));
  }
}
inline void dfs(long long u, long long from) {
  fa[u][0] = from;
  deep[u] = deep[from] + 1;
  for (long long i = 1; i <= 20; i++) {
    fa[u][i] = fa[fa[u][i - 1]][i - 1];
    w[u][i] = w[u][i - 1] + w[fa[u][i - 1]][i - 1] - f[fa[u][i - 1]];
  }
  for (long long i = head[u]; i; i = e[i].next) {
    long long v = e[i].to;
    if (v == from) continue;
    w[v][0] = f[u] + f[v] - e[i].w;
    if (f[v] - (e[i].w << 1) > 0) w[v][0] -= f[v] - (e[i].w << 1);
    g[v] = max(g[v], g[u] + w[v][0] - e[i].w - f[v]);
    dfs(v, u);
  }
}
inline long long query(long long u, long long v) {
  long long x = u, y = v, ans = 0;
  if (deep[u] < deep[v]) swap(u, v);
  for (long long i = 20; i >= 0; i--)
    if (deep[fa[u][i]] >= deep[v]) {
      ans += w[u][i] - f[u];
      u = fa[u][i];
    }
  if (u == v) return ans + (deep[x] > deep[y] ? f[x] : f[y]) + g[u];
  for (long long i = 20; i >= 0; i--)
    if (fa[u][i] != fa[v][i]) {
      ans += w[u][i] - f[u] + w[v][i] - f[v];
      u = fa[u][i], v = fa[v][i];
    }
  return ans + w[u][0] - f[u] + w[v][0] - f[v] + f[x] + f[y] + g[fa[u][0]] -
         f[fa[u][0]];
}
signed main() {
  scanf("%lld%lld", &n, &Q);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1, x, y, z; i < n; i++) {
    scanf("%lld%lld%lld", &x, &y, &z);
    add(x, y, z);
    add(y, x, z);
  }
  dp(1, 0);
  dfs(1, 0);
  while (Q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    printf("%lld\n", query(x, y));
  }
  return 0;
}