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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rand(int l, int r) {
  uniform_int_distribution<int> uid(l, r);
  return uid(rng);
}
const int N = 3e5 + 5;
const int LOGN = 19;
vector<int> g[N];
int U[N], V[N], W[N];
int dp[LOGN][N], depth[N];
long long a[N], d[N], f[N], pp[N], er[N], pre[N];
int lca(int u, int v) {
  if (depth[u] < depth[v]) swap(u, v);
  for (int k = LOGN - 1; k >= 0; k--) {
    if (depth[dp[k][u]] >= depth[v]) u = dp[k][u];
  }
  if (u == v) return u;
  for (int k = LOGN - 1; k >= 0; k--) {
    if (dp[k][u] != dp[k][v]) u = dp[k][u], v = dp[k][v];
  }
  return dp[0][u];
}
int go_up(int u, int d) {
  for (int k = LOGN - 1; k >= 0; k--) {
    if ((d >> k) & 1) {
      u = dp[k][u];
    }
  }
  return u;
}
void dfs(int u, int p) {
  depth[u] = depth[p] + 1;
  dp[0][u] = p;
  d[u] = a[u];
  for (int id : g[u]) {
    int v = U[id] ^ V[id] ^ u;
    if (v == p) continue;
    pre[v] = pre[u] - W[id];
    dfs(v, u);
    er[v] = max(0LL, d[v] - 2 * W[id]);
    d[u] += er[v];
  }
  for (int id : g[u]) {
    int v = U[id] ^ V[id] ^ u;
    if (v == p) continue;
    f[v] = d[u] - er[v];
  }
}
void dfs1(int u, int p) {
  for (int id : g[u]) {
    int v = U[id] ^ V[id] ^ u;
    if (v == p) continue;
    pp[v] = max(0LL, pp[u] + f[v] - 2 * W[id]);
    f[v] += f[u];
    dfs1(v, u);
  }
}
long long query(int u, int v) {
  int lc = lca(u, v);
  long long ans = d[lc] + pp[lc];
  if (u != lc) {
    int x = go_up(u, depth[u] - depth[lc] - 1);
    ans -= er[x];
    ans += f[u] - f[x];
    ans += d[u];
  }
  if (v != lc) {
    int x = go_up(v, depth[v] - depth[lc] - 1);
    ans -= er[x];
    ans += f[v] - f[x];
    ans += d[v];
  }
  ans += pre[u] + pre[v] - 2 * pre[lc];
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n - 1; i++) {
    cin >> U[i] >> V[i] >> W[i];
    g[U[i]].push_back(i);
    g[V[i]].push_back(i);
  }
  dfs(1, 0);
  dfs1(1, 0);
  for (int k = 1; k <= LOGN - 1; k++) {
    for (int i = 1; i <= n; i++) {
      dp[k][i] = dp[k - 1][dp[k - 1][i]];
    }
  }
  while (q--) {
    int u, v;
    cin >> u >> v;
    cout << query(u, v) << '\n';
  }
  return 0;
}