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;
long long a[300001], s[300001];
struct edge {
  unsigned p, u, v;
  long long w, s;
} es[300001 << 1];
unsigned eh[300001], ec;
void adde(unsigned u, unsigned v, long long w) {
  es[ec] = {eh[u], u, v, w, 0};
  eh[u] = ec++;
  es[ec] = {eh[v], v, u, w, 0};
  eh[v] = ec++;
}
void dfs1(unsigned u, unsigned f) {
  for (unsigned ei = eh[u]; ~ei; ei = es[ei].p) {
    edge& e = es[ei];
    if (e.v == f) continue;
    dfs1(e.v, u);
    e.s = max(0ll, s[e.v] + a[e.v] - 2 * e.w);
    s[u] += e.s;
  }
}
void dfs2(unsigned u, unsigned f) {
  for (unsigned ei = eh[u]; ~ei; ei = es[ei].p) {
    edge e = es[ei];
    if (e.v == f) continue;
    es[ei ^ 1].s = max(0ll, s[u] - e.s + a[u] - 2 * e.w);
    s[e.v] += es[ei ^ 1].s;
    dfs2(e.v, u);
  }
}
unsigned d[300001], mp[300001][20];
long long ms[300001][20];
void dfs3(unsigned u, unsigned f, long long k) {
  d[u] = d[f] + 1;
  mp[u][0] = f;
  ms[u][0] = k;
  for (unsigned ei = eh[u]; ~ei; ei = es[ei].p) {
    edge e = es[ei];
    if (e.v == f) continue;
    dfs3(e.v, u, s[e.v] - e.s - es[ei ^ 1].s + a[e.v] - e.w);
  }
}
long long lca(unsigned u, unsigned v) {
  long long res = 0;
  if (d[u] < d[v]) swap(u, v);
  for (unsigned i = 19; ~i; --i)
    if ((d[u] - d[v] >> i) & 1) {
      res += ms[u][i];
      u = mp[u][i];
    }
  if (u == v) return res + s[u] + a[u];
  for (unsigned i = 19; ~i; --i)
    if (mp[u][i] != mp[v][i]) {
      res += ms[u][i] + ms[v][i];
      u = mp[u][i];
      v = mp[v][i];
    }
  return res + s[mp[u][0]] + a[mp[u][0]] + ms[u][0] + ms[v][0];
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  unsigned n, q;
  cin >> n >> q;
  for (unsigned i = 1; i <= n; ++i) cin >> a[i];
  memset(eh, 0xff, sizeof(eh));
  for (unsigned i = 1; i < n; ++i) {
    unsigned u, v;
    long long w;
    cin >> u >> v >> w;
    adde(u, v, w);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  dfs3(1, 0, 0);
  for (unsigned j = 1; j != 20; ++j)
    for (unsigned i = 1; i <= n; ++i) {
      mp[i][j] = mp[mp[i][j - 1]][j - 1];
      ms[i][j] = ms[i][j - 1] + ms[mp[i][j - 1]][j - 1];
    }
  while (q--) {
    unsigned u, v;
    cin >> u >> v;
    cout << lca(u, v) << endl;
  }
  return 0;
}