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 double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int Minf = 0x7f7f7f7f;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const long long MINF = 0x7f7f7f7f7f7f7f7fll;
const int N = 3e5 + 50;
int head[N], to[N << 1], nxt[N << 1], val[N << 1], tot;
int fa[N][21], dep[N], a[N];
long long g[N], h[N], up[N], k[N];
void add(int x, int y, int z) {
  to[++tot] = y;
  nxt[tot] = head[x];
  head[x] = tot;
  val[tot] = z;
}
void dfs(int now, int Fa) {
  dep[now] = dep[Fa] + 1;
  g[now] = a[now];
  fa[now][0] = Fa;
  for (int i = (1); i <= 20; ++i) fa[now][i] = fa[fa[now][i - 1]][i - 1];
  for (int i = head[now]; i; i = nxt[i]) {
    if (to[i] == Fa) continue;
    dfs(to[i], now);
    h[to[i]] = max(g[to[i]] - 2 * val[i], 0ll);
    g[now] += h[to[i]];
    k[to[i]] = val[i];
  }
}
void dfs2(int now, int Fa) {
  if (Fa) up[now] = max(up[Fa] + g[Fa] - h[now] - 2 * k[now], 0ll);
  for (int i = head[now]; i; i = nxt[i]) {
    if (to[i] == Fa) continue;
    dfs2(to[i], now);
  }
}
void dfs3(int now, int Fa) {
  g[now] += g[Fa];
  h[now] += h[Fa];
  k[now] += k[Fa];
  for (int i = head[now]; i; i = nxt[i]) {
    if (to[i] == Fa) continue;
    dfs3(to[i], now);
  }
}
int getlca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = (20); i >= 0; --i)
    if (dep[fa[x][i]] >= dep[y]) x = fa[x][i];
  if (x == y) return x;
  for (int i = (20); i >= 0; --i)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
int kfa(int x, int k) {
  for (int i = (0); i <= 20; ++i)
    if (k & (1 << i)) x = fa[x][i];
  return x;
}
int main() {
  int n, q, x, y, z;
  scanf("%d%d", &n, &q);
  for (int i = (1); i <= n; ++i) scanf("%d", &a[i]);
  for (int i = (2); i <= n; ++i)
    scanf("%d%d%d", &x, &y, &z), add(x, y, z), add(y, x, z);
  dfs(1, 0);
  dfs2(1, 0);
  dfs3(1, 0);
  for (int i = (1); i <= q; ++i) {
    scanf("%d%d", &x, &y);
    int lca = getlca(x, y);
    if (lca == y || lca == y) {
      if (lca == x) swap(x, y);
      long long ans = g[x] - g[fa[y][0]] - h[x] + h[y] + up[y] - k[x] + k[y];
      printf("%lld\n", ans);
    } else {
      long long ans = g[x] - g[lca] + g[y] - g[fa[lca][0]] - h[x] + h[lca] -
                      h[y] + h[lca] + up[lca] - k[x] + k[lca] - k[y] + k[lca];
      printf("%lld\n", ans);
    }
  }
  return 0;
}