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;
constexpr int N = 3e5 + 10;
constexpr int LG = 19;
constexpr int MOD = 1e9 + 7;
long long n, q, a[N], dp[N], dp2[N], u, v, w, h[N], par2[LG][N];
long long par[LG][N], lg[N], distpar[N], cost[LG][N], weight[N];
vector<pair<long long, long long>> node[N];
void dfs(int v, int p = -1) {
  dp[v] = a[v];
  par2[0][v] = v;
  for (auto i : node[v]) {
    if (i.first != p) {
      weight[i.first] = weight[v] + i.second;
      par[0][i.first] = v;
      h[i.first] = h[v] + 1;
      distpar[i.first] = i.second;
      dfs(i.first, v);
      if (2 * i.second < dp[i.first]) dp[v] += dp[i.first] - 2 * i.second;
    }
  }
  cost[0][v] = dp[v];
}
void dfs2(int v, int p = -1) {
  long long others = dp[v] + dp2[v], now = 0;
  for (auto i : node[v]) {
    if (i.first != p) {
      now = others;
      if (2 * i.second < dp[i.first]) now += 2 * i.second - dp[i.first];
      if (now > 2 * i.second) dp2[i.first] = now - 2 * i.second;
      dfs2(i.first, v);
    }
  }
}
int lca(int a, int b) {
  if (h[a] < h[b]) swap(a, b);
  while (h[a] != h[b]) a = par[lg[h[a] - h[b]]][a];
  for (int i = LG - 1; ~i; i--) {
    if (par[i][a] != par[i][b]) {
      a = par[i][a];
      b = par[i][b];
    }
  }
  if (a == b) return a;
  return par[0][a];
}
int find_par(int a, int diff) {
  while (diff) a = par[lg[diff]][a], diff -= 1 << lg[diff];
  return a;
}
long long get(int a, int diff) {
  long long ret = 0;
  int z = 0, newdiff = 0;
  while (diff) {
    z = par2[lg[diff]][a];
    ret += cost[lg[diff]][a];
    newdiff = diff - (1 << lg[diff]);
    if (newdiff && dp[z] - 2 * distpar[z] > 0) ret -= dp[z] - 2 * distpar[z];
    a = par[lg[diff]][a];
    diff = newdiff;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    cin >> u >> v >> w;
    node[u].push_back({v, w});
    node[v].push_back({u, w});
  }
  par[0][1] = 1;
  dfs(1);
  dfs2(1);
  for (int i = 2; i < N; i++) lg[i] = lg[i >> 1] + 1;
  for (int i = 1; i < LG; i++) {
    for (int j = 1; j <= n; j++) {
      par[i][j] = par[i - 1][par[i - 1][j]];
      par2[i][j] = par[i - 1][par2[i - 1][j]];
    }
  }
  for (int i = 1; i < LG; i++) {
    for (int j = 1; j <= n; j++) {
      cost[i][j] = cost[i - 1][j] + cost[i - 1][par[i - 1][j]];
      v = par2[i - 1][j];
      if (dp[v] - 2 * distpar[v] > 0) cost[i][j] -= dp[v] - 2 * distpar[v];
    }
  }
  while (q--) {
    cin >> u >> v;
    if (u == v) {
      cout << dp[v] + dp2[v] << '\n';
      continue;
    }
    int mylca = lca(u, v);
    long long ans =
        2 * weight[mylca] - weight[u] - weight[v] + dp2[mylca] + dp[mylca];
    if (u == mylca) {
      ans += get(v, h[v] - h[mylca]);
      int zv = find_par(v, h[v] - h[mylca] - 1);
      if (dp[zv] - 2 * distpar[zv] > 0) ans -= dp[zv] - 2 * distpar[zv];
    } else if (v == mylca) {
      ans += get(u, h[u] - h[mylca]);
      int zu = find_par(u, h[u] - h[mylca] - 1);
      if (dp[zu] - 2 * distpar[zu] > 0) ans -= dp[zu] - 2 * distpar[zu];
    } else {
      ans += get(v, h[v] - h[mylca]) + get(u, h[u] - h[mylca]);
      int zu = find_par(u, h[u] - h[mylca] - 1);
      int zv = find_par(v, h[v] - h[mylca] - 1);
      if (dp[zu] - 2 * distpar[zu] > 0) ans -= dp[zu] - 2 * distpar[zu];
      if (dp[zv] - 2 * distpar[zv] > 0) ans -= dp[zv] - 2 * distpar[zv];
    }
    cout << ans << '\n';
  }
  return 0;
}