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 v, w;
  Edge* next;
};
const int N = 400010;
int n, q;
long long sub[N], lk[N], spec[N], asum[N], wsum[N];
int a[N], f[N], prt[N], prtw[N], qu[N], qv[N], lca[N], ord[N];
Edge* g[N];
vector<pair<int, int> > qry[N];
int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
void adde(int u, int v, int w);
void tarjan(int u);
void dfs(int u);
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int rep = 1; rep < n; ++rep) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    adde(u, v, w);
    adde(v, u, w);
  }
  for (int i = 1; i <= q; ++i) {
    scanf("%d%d", &qu[i], &qv[i]);
    qry[qu[i]].emplace_back(qv[i], i);
    qry[qv[i]].emplace_back(qu[i], i);
  }
  tarjan(1);
  dfs(1);
  for (int i = 2; i <= n; ++i) {
    int u = ord[i];
    spec[u] =
        max(0LL, spec[prt[u]] + a[prt[u]] - 2 * prtw[u] + sub[prt[u]] - lk[u]);
  }
  copy(a + 1, a + n + 1, asum + 1);
  copy(prtw + 1, prtw + n + 1, wsum + 1);
  for (int i = 2; i <= n; ++i) {
    int u = ord[i];
    sub[u] += sub[prt[u]];
    lk[u] += lk[prt[u]];
    asum[u] += asum[prt[u]];
    wsum[u] += wsum[prt[u]];
  }
  for (int i = 1; i <= q; ++i) {
    int u = qu[i], v = qv[i], l = lca[i];
    printf("%lld\n", sub[u] - sub[prt[l]] + sub[v] - sub[l] -
                         (lk[u] + lk[v] - 2 * lk[l]) + spec[l] +
                         (asum[u] - asum[prt[l]] + asum[v] - asum[l]) -
                         (wsum[u] + wsum[v] - 2 * wsum[l]));
  }
  return 0;
}
void dfs(int u) {
  static int t;
  ord[++t] = u;
  for (Edge* p = g[u]; p; p = p->next)
    if (prt[p->v] == u) {
      dfs(p->v);
      lk[p->v] = max(0LL, sub[p->v] + a[p->v] - 2 * p->w);
      sub[u] += lk[p->v];
    }
}
void tarjan(int u) {
  f[u] = u;
  for (Edge* p = g[u]; p; p = p->next)
    if (!f[p->v]) {
      prt[p->v] = u;
      prtw[p->v] = p->w;
      tarjan(p->v);
      f[p->v] = u;
    }
  for (auto q : qry[u])
    if (f[q.first]) lca[q.second] = find(q.first);
}
void adde(int u, int v, int w) {
  static Edge pool[N * 2];
  static Edge* p = pool;
  p->v = v;
  p->w = w;
  p->next = g[u];
  g[u] = p;
  ++p;
}