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 n, q;
struct edge {
  long long v, w, next;
} e[300010 << 1];
long long eid, p[300010];
void insert(long long u, long long v, long long w) {
  e[++eid] = {v, w, p[u]};
  p[u] = eid;
}
long long w[300010], ed[300010], dep[300010];
long long f[300010][20], c1[300010][20], c2[300010][20];
void dfs1(long long u, long long fa, long long x) {
  ed[u] = x;
  f[u][0] = fa;
  dep[u] = dep[fa] + 1;
  c1[u][0] = w[u] - ed[u];
  for (long long i = 1; (1 << i) <= dep[u]; ++i) {
    f[u][i] = f[f[u][i - 1]][i - 1];
    c1[u][i] = c1[u][i - 1] + c1[f[u][i - 1]][i - 1];
  }
  for (long long i = p[u]; i; i = e[i].next) {
    long long v = e[i].v;
    if (fa == v) continue;
    dfs1(v, u, e[i].w);
  }
}
long long dp[3][300010], vis[300010];
void dfs2(long long u, long long f) {
  dp[1][u] = 0;
  for (long long i = p[u]; i; i = e[i].next) {
    long long v = e[i].v;
    if (f == v) continue;
    dfs2(v, u);
    if (w[v] + dp[1][v] > 2 * ed[v])
      dp[1][u] += w[v] + dp[1][v] - 2 * ed[v], vis[v] = 1;
  }
}
void dfs3(long long u, long long fa) {
  long long a = max(0ll, dp[2][fa]);
  long long b = max(0ll, dp[1][fa] - vis[u] * (w[u] + dp[1][u] - 2 * ed[u]));
  dp[2][u] = max(0ll, w[fa] - 2 * ed[u] + a + b);
  c2[u][0] = dp[1][u] - vis[u] * (dp[1][u] + w[u] - 2 * ed[u]);
  for (long long i = 1; (1 << i) <= dep[u]; ++i)
    c2[u][i] = c2[u][i - 1] + c2[f[u][i - 1]][i - 1];
  for (long long i = p[u]; i; i = e[i].next) {
    long long v = e[i].v;
    if (fa == v) continue;
    dfs3(v, u);
  }
}
long long lca(long long u, long long v) {
  if (dep[u] < dep[v]) swap(u, v);
  long long x = 0, ans = 0, lca = 0, z = u;
  for (x = 0; (1 << x) <= dep[u]; ++x)
    ;
  --x;
  for (long long i = x; i >= 0; --i)
    if (dep[f[u][i]] >= dep[v]) ans += c1[u][i] + c2[u][i], u = f[u][i];
  for (long long i = x; i >= 0; --i)
    if (dep[f[z][i]] > dep[v]) z = f[z][i];
  if (u == v) return dp[2][u] + ans + w[u] + dp[1][u];
  for (long long i = x; i >= 0; --i)
    if (f[u][i] != f[v][i]) {
      ans += c1[u][i] + c1[v][i] + c2[u][i] + c2[v][i];
      u = f[u][i], v = f[v][i];
    }
  lca = f[u][0];
  return ans + w[lca] + dp[1][lca] + dp[2][lca] + dp[1][u] + dp[1][v] +
         c1[u][0] + c1[v][0] - vis[u] * (vis[u] * dp[1][u] + w[u] - 2 * ed[u]) -
         vis[v] * (vis[v] * dp[1][v] + w[v] - 2 * ed[v]);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> q;
  for (long long i = 1; i <= n; ++i) cin >> w[i];
  for (long long i = 1; i < n; ++i) {
    long long u, v, w;
    cin >> u >> v >> w;
    insert(u, v, w);
    insert(v, u, w);
  }
  dfs1(1, 0, 0);
  dfs2(1, 0);
  dfs3(1, 0);
  while (q--) {
    long long u, v;
    cin >> u >> v;
    cout << lca(u, v) << endl;
  }
  return 0;
}