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 {
  int to, next;
  long long len;
} edge[300010 << 1];
int num, head[300010], fa[300010][25], n, m, d[300010], val[300010];
long long f[300010], g[300010], dp[300010][25];
inline int read() {
  int s = 0, w = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') w = -1;
  for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
  return s * w;
}
void addedge(int x, int y, int z) {
  edge[++num] = (Edge){y, head[x], z}, head[x] = num;
}
void calc(int u, int pre) {
  f[u] = val[u];
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v != pre) {
      calc(v, u);
      if (f[v] - 2LL * edge[i].len > 0LL) f[u] += f[v] - 2LL * edge[i].len;
    }
  }
}
void build(int u, int pre) {
  d[u] = d[pre] + 1, fa[u][0] = pre;
  for (int i = 0; fa[u][i]; ++i) {
    fa[u][i + 1] = fa[fa[u][i]][i];
    dp[u][i + 1] = dp[u][i] + dp[fa[u][i]][i] - f[fa[u][i]];
  }
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v != pre) {
      dp[v][0] = f[v] + f[u] - edge[i].len;
      if (f[v] - 2LL * edge[i].len > 0LL) dp[v][0] -= f[v] - 2LL * edge[i].len;
      g[v] = max(g[v], g[u] + dp[v][0] - f[v] - edge[i].len);
      build(v, u);
    }
  }
}
long long lca(int u, int v) {
  int x = u, y = v;
  if (d[u] < d[v]) swap(u, v);
  long long ans = 0;
  for (int i = 20; i >= 0; --i)
    if (d[u] - (1 << i) >= d[v]) {
      ans += dp[u][i] - f[u], u = fa[u][i];
    }
  if (u == v) return ans + g[u] + (d[x] > d[y] ? f[x] : f[y]);
  for (int i = 20; i >= 0; --i)
    if (fa[u][i] != fa[v][i]) {
      ans += dp[u][i] + dp[v][i] - f[u] - f[v];
      u = fa[u][i], v = fa[v][i];
    }
  return ans + dp[u][0] + dp[v][0] - f[u] - f[v] - f[fa[u][0]] + g[fa[u][0]] +
         f[x] + f[y];
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; ++i) val[i] = read();
  for (int i = 1; i < n; ++i) {
    int x = read(), y = read();
    long long z = read();
    addedge(x, y, z), addedge(y, x, z);
  }
  calc(1, 0);
  build(1, 0);
  while (m--) {
    int x = read(), y = read();
    printf("%lld\n", lca(x, y));
  }
  return 0;
}