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 read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int Size = 400005;
struct Edge {
  long long pos, v;
  Edge(long long a, long long b) {
    pos = a;
    v = b;
  }
};
vector<Edge> to[Size];
long long n, q, a[Size], deep[Size], d[Size];
long long anc[Size][21], toroot[Size];
long long qu[Size], qv[Size];
long long dp1[Size], dp2[Size], dp[Size][21];
void dfs1(long long x, long long fx) {
  deep[x] = deep[fx] + 1;
  anc[x][0] = fx;
  for (int i = 1; i <= 20; i++) {
    anc[x][i] = anc[anc[x][i - 1]][i - 1];
  }
  int len = to[x].size();
  for (int i = 0; i < len; i++) {
    int nxt = to[x][i].pos;
    if (nxt != fx) {
      d[nxt] = d[x] + a[nxt] - to[x][i].v;
      dfs1(nxt, x);
      dp1[x] += max(dp1[nxt] - (to[x][i].v << 1), 0ll);
      toroot[nxt] = to[x][i].v;
    }
  }
}
void dfs2(long long x, long long fx) {
  dp[x][0] = dp1[fx] - max(dp1[x] - (toroot[x] << 1), 0ll) - a[fx];
  dp2[x] = max(dp2[fx] + dp[x][0] + a[fx] - (toroot[x] << 1), 0ll);
  for (int i = 1; i <= 20; i++) {
    dp[x][i] = dp[x][i - 1] + dp[anc[x][i - 1]][i - 1];
  }
  int len = to[x].size();
  for (int i = 0; i < len; i++) {
    long long nxt = to[x][i].pos;
    if (nxt != fx) {
      dfs2(nxt, x);
    }
  }
}
void init() {
  n = read();
  q = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    dp1[i] = a[i];
  }
  for (int i = 1; i < n; i++) {
    long long u = read();
    long long v = read();
    long long w = read();
    to[u].push_back(Edge(v, w));
    to[v].push_back(Edge(u, w));
  }
  for (int i = 1; i <= q; i++) {
    qu[i] = read();
    qv[i] = read();
  }
  dfs1(1, 0);
  dfs2(1, 0);
}
int lca(long long u, long long v) {
  if (deep[u] < deep[v]) swap(u, v);
  for (int i = 20; i >= 0; i--) {
    if (deep[anc[u][i]] >= deep[v]) {
      u = anc[u][i];
    }
  }
  if (u == v) return u;
  for (int i = 20; i >= 0; i--) {
    if (anc[u][i] != anc[v][i]) {
      u = anc[u][i];
      v = anc[v][i];
    }
  }
  return anc[u][0];
}
long long Query(int u, int v) {
  if (u == v) {
    return dp1[u] + dp2[u];
  }
  if (deep[u] < deep[v]) swap(u, v);
  long long nu = u, nv = v;
  long long ancest = lca(u, v), ans = 0;
  if (ancest == v) {
    for (int i = 20; i >= 0; i--) {
      if (deep[anc[u][i]] >= deep[ancest]) {
        ans += dp[u][i];
        u = anc[u][i];
      }
    }
    long long dist = d[nu] - d[v];
    ans += dp1[nu] + a[v] - a[nu] + dp2[v] + dist;
    return ans;
  }
  for (int i = 20; i >= 0; i--) {
    if (deep[anc[u][i]] > deep[ancest]) {
      ans += dp[u][i];
      u = anc[u][i];
    }
    if (deep[anc[v][i]] > deep[ancest]) {
      ans += dp[v][i];
      v = anc[v][i];
    }
  }
  long long s1 = dp1[nu] - a[nu], s2 = dp1[nv] - a[nv],
            s3 = dp2[ancest] + dp1[ancest] - a[ancest];
  long long mu = max(dp1[u] - (toroot[u] << 1), 0ll);
  long long mv = max(dp1[v] - (toroot[v] << 1), 0ll);
  long long dist1 = d[nu] - d[anc[u][0]], dist2 = d[nv] - d[anc[v][0]];
  ans += a[ancest] + s1 + s2 + s3 - mu - mv + dist1 + dist2;
  return ans;
}
int main() {
  init();
  for (int i = 1; i <= q; i++) {
    long long ans = Query(qu[i], qv[i]);
    printf("%I64d\n", ans);
  }
  return 0;
}