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 MAXN = 3e5 + 10;
long long N, M, cnt = -1, f[MAXN], g[MAXN], dep[MAXN], fa[MAXN][21],
                st[MAXN][21], head[MAXN], val[MAXN];
struct edge {
  long long nt, to, v;
} E[MAXN << 1];
void add(long long, long long, long long);
void pre(long long, long long);
void dfs(long long, long long);
long long calc(long long, long long);
int main() {
  memset(head, -1, sizeof(head));
  scanf("%lld%lld", &N, &M);
  for (long long i = 1; i <= N; i++) scanf("%lld", val + i);
  for (long long i = 1, x, y, v; i < N; i++) {
    scanf("%lld%lld%lld", &x, &y, &v);
    add(x, y, v), add(y, x, v);
  }
  pre(1, 0);
  dfs(1, 0);
  for (long long i = 1, x, y; i <= M; i++)
    scanf("%lld%lld", &x, &y), printf("%lld\n", calc(x, y));
  return 0;
}
long long calc(long long x, long long y) {
  if (dep[x] < dep[y]) swap(x, y);
  long long ret = 0, tem = f[x], dis = dep[x] - dep[y];
  for (long long i = 20; ~i; i--)
    if ((dis >> i) & 1) ret += st[x][i] - f[x], x = fa[x][i];
  if (x == y) return ret + tem + g[x];
  tem += f[y];
  for (long long i = 20; ~i; i--)
    if (fa[x][i] != fa[y][i])
      ret += st[x][i] - f[x], ret += st[y][i] - f[y], x = fa[x][i],
                                                      y = fa[y][i];
  return ret + st[x][0] + st[y][0] - f[x] - f[y] - f[fa[x][0]] + g[fa[x][0]] +
         tem;
}
void dfs(long long n, long long ff) {
  dep[n] = dep[ff] + 1, fa[n][0] = ff;
  for (long long i = 1; i <= 20; i++) {
    fa[n][i] = fa[fa[n][i - 1]][i - 1];
    st[n][i] = st[fa[n][i - 1]][i - 1] - f[fa[n][i - 1]] + st[n][i - 1];
  }
  for (long long i = head[n]; ~i; i = E[i].nt) {
    long long v = E[i].to;
    if (v == ff) continue;
    st[v][0] = f[n] + f[v] - E[i].v;
    if (f[v] - 2 * E[i].v > 0) st[v][0] -= f[v] - 2 * E[i].v;
    g[v] = max(g[v], g[n] + st[v][0] - E[i].v - f[v]);
    dfs(v, n);
  }
}
void pre(long long n, long long ff) {
  f[n] = val[n];
  for (long long i = head[n]; ~i; i = E[i].nt) {
    long long v = E[i].to;
    if (v == ff) continue;
    pre(v, n);
    f[n] += max(0LL, f[v] - E[i].v * 2);
  }
}
void add(long long x, long long y, long long v) {
  cnt++;
  E[cnt].to = y;
  E[cnt].nt = head[x];
  head[x] = cnt;
  E[cnt].v = v;
}