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 power(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y / 2, x = (x * x) % 1000000007;
  }
  return res % 1000000007;
}
long long child[3 * 100001];
long long parent[3 * 100001];
long long val[3 * 100001];
vector<vector<pair<long long, long long> > > arr(3 * 100001);
long long lca[300001][20];
long long depth[300001];
long long n, q;
void dfs0(long long u, long long par) {
  child[u] = val[u];
  long long i;
  for (i = 0; i < arr[u].size(); i++) {
    long long v = arr[u][i].first;
    long long wt = arr[u][i].second;
    if (v != par) {
      dfs0(v, u);
      child[u] += max(child[v] - 2LL * wt, 0LL);
    }
  }
}
void dfs1(long long u, long long par, long long wt) {
  if (par != 0)
    parent[u] =
        parent[par] + child[par] - max(child[u] - 2LL * wt, 0LL) - 2LL * wt;
  parent[u] = max(parent[u], 0LL);
  long long i;
  for (i = 0; i < arr[u].size(); i++) {
    long long v = arr[u][i].first;
    if (v != par) dfs1(v, u, arr[u][i].second);
  }
}
void dfslca(long long u, long long v, long long d) {
  lca[u][0] = v;
  depth[u] = d;
  long long i;
  for (i = 0; i < arr[u].size(); i++) {
    if (arr[u][i].first != v) dfslca(arr[u][i].first, u, d + 1);
  }
}
void preprolca() {
  dfslca(1, 0, 1);
  long long i, j;
  for (j = 1; j < 20; j++)
    for (i = 1; i <= 300000; i++) lca[i][j] = lca[lca[i][j - 1]][j - 1];
}
long long query(long long u, long long v) {
  if (depth[u] > depth[v]) swap(u, v);
  if (u == v) return u;
  long long i;
  for (i = 19; i >= 0; i--) {
    if (depth[u] + (1 << i) <= depth[v]) v = lca[v][i];
  }
  if (u == v) return u;
  for (i = 19; i >= 0; i--) {
    if (lca[u][i] != lca[v][i]) u = lca[u][i], v = lca[v][i];
  }
  return lca[u][0];
}
vector<vector<pair<long long, long long> > > hola(4 * 100001);
long long ans[4 * 100001];
long long temp[3 * 100001];
long long seg[1200004];
void update(long long low, long long high, long long index, long long pos,
            long long val) {
  if (low > high or low > pos or high < pos) return;
  if (low == high) {
    seg[index] += val;
    return;
  }
  long long mid = (low + high) / 2;
  update(low, mid, 2 * index, pos, val);
  update(mid + 1, high, 2 * index + 1, pos, val);
  seg[index] = seg[2 * index] + seg[2 * index + 1];
}
long long answerdo(long long low, long long high, long long index, long long l,
                   long long r) {
  if (low > r or low > high or high < l) return 0;
  if (low >= l and high <= r) return seg[index];
  long long mid = (low + high) / 2;
  return answerdo(low, mid, 2 * index, l, r) +
         answerdo(mid + 1, high, 2 * index + 1, l, r);
}
void dfs(long long u, long long par) {
  long long i;
  update(1, n, 1, depth[u], child[u]);
  for (i = 0; i < hola[u].size(); i++) {
    long long v = hola[u][i].first;
    long long q = hola[u][i].second;
    long long va = answerdo(1, n, 1, depth[v], depth[u]);
    ans[q] += (va);
  }
  for (i = 0; i < arr[u].size(); i++) {
    long long v = arr[u][i].first;
    long long wt = arr[u][i].second;
    if (v != par) {
      temp[u] = max(child[v] - 2LL * wt, 0LL);
      update(1, n, 1, depth[u], (temp[u] + wt) * -1);
      dfs(v, u);
      update(1, n, 1, depth[u], temp[u] + wt);
      temp[u] = 0;
    }
  }
  update(1, n, 1, depth[u], -1 * child[u]);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  long long i;
  for (i = 1; i <= n; i++) cin >> val[i];
  for (i = 0; i < n - 1; i++) {
    long long u, v, wt;
    cin >> u >> v >> wt;
    arr[u].push_back(make_pair(v, wt));
    arr[v].push_back(make_pair(u, wt));
  }
  dfs0(1, 0);
  dfs1(1, 0, 0);
  preprolca();
  for (i = 0; i < q; i++) {
    long long u, v;
    cin >> u >> v;
    if (v == u) {
      ans[i] = child[u] + parent[u];
      continue;
    }
    long long l = query(u, v);
    if (v != u) {
      hola[u].push_back(make_pair(l, i));
      hola[v].push_back(make_pair(l, i));
    } else
      hola[u].push_back(make_pair(l, i));
    ans[i] += parent[l];
    ans[i] -= (child[l]);
  }
  dfs(1, 0);
  for (i = 0; i < q; i++) cout << ans[i] << "\n";
}