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;
int n, q;
int a[300010];
vector<pair<int, int> > g[300010];
int tin[300010], tout[300010], dfstimer;
int p[20][300010];
long long dfull[300010], dsubtree[300010];
long long topar[300010];
vector<pair<int, int> > b[300100];
long long an[400010];
int h[300010];
long long fen[300010];
void upd(int pos, long long val) {
  for (; pos < n; pos |= (pos + 1)) fen[pos] += val;
}
long long getsum(int r) {
  long long result = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) result += fen[r];
  return result;
}
long long getsum(int l, int r) { return getsum(r) - getsum(l - 1); }
void dfs1(int v, int par, int edge, int he) {
  h[v] = he;
  tin[v] = dfstimer++;
  p[0][v] = par;
  for (int i = 1; i < 20; i++) {
    p[i][v] = p[i - 1][p[i - 1][v]];
  }
  dsubtree[v] = a[v];
  for (auto e : g[v]) {
    int to = e.first;
    int len = e.second;
    if (to != par) {
      dfs1(to, v, e.second, he + 1);
      dsubtree[v] += max(0ll, dsubtree[to] - 2 * e.second);
    }
  }
  tout[v] = dfstimer++;
}
void dfs2(int v, int par, int edge) {
  if (v != par) {
    long long value = dfull[par];
    value -= max(0ll, dsubtree[v] - 2 * edge);
    dfull[v] = dsubtree[v] + max(0ll, value - 2 * edge);
    topar[v] = max(0ll, value - 2 * edge);
  } else
    dfull[v] = dsubtree[v];
  for (auto e : g[v]) {
    int to = e.first;
    if (to != par) {
      dfs2(to, v, e.second);
    }
  }
}
bool upper(int a, int b) { return tin[a] <= tin[b] && tout[b] <= tout[a]; }
int lca(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  for (int i = 19; i >= 0; i--) {
    if (!upper(p[i][a], b)) a = p[i][a];
  }
  return p[0][a];
}
int get(int a, int b) {
  for (int i = 19; i >= 0; i--) {
    if (!upper(p[i][a], b)) a = p[i][a];
  }
  return a;
}
int getedge(int a, int b) {
  int lo = 0, hi = g[a].size() - 1;
  while (lo != hi) {
    int mi = (lo + hi) >> 1;
    if (g[a][mi].first < b)
      lo = mi + 1;
    else
      hi = mi;
  }
  return g[a][lo].second;
}
void dfs3(int v, int par) {
  upd(h[v], dfull[v] - topar[v]);
  for (auto p : b[v]) {
    int i = p.second;
    int l = h[p.first] + 1;
    int r = h[v];
    an[i] += getsum(l, r);
  }
  for (auto e : g[v]) {
    int to = e.first;
    if (to != par) {
      upd(h[v], -(max(0ll, dsubtree[to] - 2 * e.second)) - e.second);
      dfs3(to, v);
      upd(h[v], max(0ll, dsubtree[to] - 2 * e.second) + e.second);
    }
  }
  upd(h[v], -(dfull[v] - topar[v]));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) {
    int from, to, len;
    cin >> from >> to >> len;
    g[from - 1].push_back({to - 1, len});
    g[to - 1].push_back({from - 1, len});
  }
  for (int i = 0; i < n; i++) sort(g[i].begin(), g[i].end());
  dfs1(0, 0, 0, 0);
  dfs2(0, 0, 0);
  for (int i = 0; i < q; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    if (u == v)
      an[i] = dfull[u];
    else {
      int l = lca(u, v);
      if (l != u) b[u].push_back({l, i});
      if (l != v) b[v].push_back({l, i});
      if (l == u || l == v) {
        if (u == l) swap(u, v);
        int _u = get(u, l);
        an[i] += dfull[l] - max(0ll, dsubtree[_u] - 2 * getedge(_u, l)) -
                 getedge(_u, l);
      } else {
        int _u = get(u, l);
        int _v = get(v, l);
        an[i] += dfull[l] - max(0ll, dsubtree[_u] - 2 * getedge(_u, l)) -
                 max(0ll, dsubtree[_v] - 2 * getedge(_v, l)) - getedge(_u, l) -
                 getedge(_v, l);
      }
    }
  }
  dfs3(0, 0);
  for (int i = 0; i < q; i++) cout << an[i] << '\n';
}