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 long long N = 3e5;
const long long M = 1e4;
const long long mod = 998244353;
const long long inf = 1e9;
long long read() {
  long long s = 0;
  register bool neg = 0;
  register char c = getchar();
  for (; c < '0' || c > '9'; c = getchar()) neg |= (c == '-');
  for (; c >= '0' && c <= '9'; s = s * 10 + (c ^ 48), c = getchar())
    ;
  s = (neg ? -s : s);
  return s;
}
long long a, b, head[N + 5], V[N + 5][21], F[N + 5][21], f[N + 5], g[N + 5],
    top, p[N + 5], dep[N + 5];
struct road {
  long long lst, v, w;
} s[N * 2 + 5];
void add(long long l1, long long l2, long long l3) {
  s[++top] = (road){head[l1], l2, l3};
  head[l1] = top;
}
void dfs(long long n, long long fa, long long _w) {
  dep[n] = dep[fa] + 1;
  F[n][0] = fa;
  for (long long i = (1); i <= (20); ++i) F[n][i] = F[F[n][i - 1]][i - 1];
  f[n] = p[n];
  for (long long i = head[n]; i; i = s[i].lst) {
    long long v = s[i].v, w = s[i].w;
    if (v == fa) continue;
    dfs(v, n, w);
    f[n] += max(0ll, f[v] - 2 * w);
  }
  V[n][0] = f[n] - max(0ll, f[n] - 2 * _w) - _w;
}
void Dfs(long long n, long long fa) {
  for (long long i = (1); i <= (20); ++i)
    V[n][i] = V[n][i - 1] + V[F[n][i - 1]][i - 1];
  for (long long i = head[n]; i; i = s[i].lst) {
    long long v = s[i].v, w = s[i].w;
    if (v == fa) continue;
    g[v] = max(0ll, f[n] + g[n] - max(0ll, f[v] - 2 * w) - 2 * w);
    Dfs(v, n);
  }
}
long long LCA(long long n, long long m) {
  if (dep[n] < dep[m]) swap(n, m);
  long long val = 0;
  for (long long i = (20); i >= (0); --i)
    if (dep[F[n][i]] >= dep[m]) val += V[n][i], n = F[n][i];
  if (n == m) return val + f[n] + g[n];
  for (long long i = (20); i >= (0); --i)
    if (F[n][i] != F[m][i]) {
      val += V[n][i] + V[m][i];
      n = F[n][i], m = F[m][i];
    }
  long long lca = F[n][0];
  return val + f[lca] + g[lca] + V[n][0] + V[m][0];
}
signed main() {
  a = read();
  b = read();
  for (long long i = (1); i <= (a); ++i) p[i] = read();
  long long x, y, z;
  for (long long i = (1); i <= (a - 1); ++i) {
    x = read();
    y = read();
    z = read();
    add(x, y, z);
    add(y, x, z);
  }
  dfs(1, 0, 0);
  Dfs(1, 0);
  for (long long i = (1); i <= (b); ++i) {
    x = read();
    y = read();
    printf("%lld\n", LCA(x, y));
  }
  return 0;
}