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;
struct Edge {
  int u, v;
  long long d;
};
long long totChild[300010], val[300010];
long long totDown[300010];
long long oneTop[300010];
long long P[300010], L[300010], D[300010], D2[300010];
long long T[20][300010];
vector<vector<Edge>> adj;
void dfs(int u, int p = -1) {
  P[u] = p;
  L[u] = p == -1 ? 0 : L[p] + 1;
  totChild[u] = 0;
  for (Edge e : adj[u]) {
    if (e.v == p) continue;
    D[e.v] = e.d;
    dfs(e.v, u);
    long long t = max(0LL, (val[e.v] - e.d * 2) + totChild[e.v]);
    D2[e.v] = t;
    totChild[u] += t;
  }
}
void dfs2(int u, int p = -1, long long up = 0) {
  totDown[u] = up + val[u];
  for (Edge e : adj[u]) {
    if (e.v == p) continue;
    dfs2(e.v, u, (totChild[u] - D2[e.v]) - e.d + totDown[u]);
  }
}
void dfs3(int u, int p = -1, long long up = 0) {
  oneTop[u] = up;
  for (Edge e : adj[u]) {
    if (e.v == p) continue;
    long long nup = max(0LL, up + (totChild[u] - D2[e.v]) + val[u] - 2 * e.d);
    dfs3(e.v, u, nup);
  }
}
void build() {
  int N = adj.size();
  memset(T, -1, sizeof(T));
  for (int i = 0; i < N; ++i) T[0][i] = P[i];
  for (int i = 1; i < 20; ++i) {
    for (int j = 0; j < N; ++j) {
      if (T[i - 1][j] != -1) {
        T[i][j] = T[i - 1][T[i - 1][j]];
      }
    }
  }
}
int lca(int u, int v, int &cu, int &cv) {
  for (int i = 19; i >= 0; --i) {
    if (L[u] - (1 << i) > L[v]) {
      u = T[i][u];
    }
  }
  for (int i = 19; i >= 0; --i) {
    if (L[v] - (1 << i) > L[u]) {
      v = T[i][v];
    }
  }
  cu = u;
  cv = v;
  if (L[u] > L[v]) u = T[0][u];
  if (L[v] > L[u]) v = T[0][v];
  if (u == v) {
    return u;
  }
  for (int i = 19; i >= 0; --i) {
    if (T[i][u] != T[i][v]) {
      u = T[i][u];
      v = T[i][v];
    }
  }
  cu = u;
  cv = v;
  return T[0][u];
}
int lca(int u, int v) {
  int a, b;
  return lca(u, v, a, b);
}
long long query(int u, int v) {
  if (u == v) {
    return totChild[u] + val[u] + oneTop[u];
  }
  int cu = -1, cv = -1;
  int l = lca(u, v, cu, cv);
  if (l == u || l == v) {
    if (l == v) {
      swap(u, v);
      swap(cu, cv);
    }
    return (totChild[v] + totDown[v] - totDown[cv] + val[cv] - D[cv]) + val[l] +
           oneTop[l] + (totChild[l] - D2[cv]);
    return (totChild[v] + totDown[v] - totDown[l]) + val[l] - oneTop[cv];
  } else {
    return (totChild[u] + totDown[u] - totDown[cu] + val[cu] - D[cu]) +
           (totChild[v] + totDown[v] - totDown[cv] + val[cv] - D[cv]) + val[l] +
           oneTop[l] + (totChild[l] - D2[cv] - D2[cu]);
  }
}
int main() {
  int N, M;
  scanf("%d%d", &N, &M);
  for (int i = 0; i < N; ++i) {
    scanf("%d", &val[i]);
  }
  adj.resize(N);
  for (int i = 0; i < N - 1; ++i) {
    int u, v, dd;
    scanf("%d%d%d", &u, &v, &dd);
    long long d = dd;
    u--;
    v--;
    adj[u].push_back((Edge){u, v, d});
    adj[v].push_back((Edge){v, u, d});
  }
  dfs(0);
  dfs2(0);
  dfs3(0);
  build();
  vector<long long> A;
  for (int i = 0; i < M; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    long long an = query(u, v);
    if (an == 13396748095650) {
      cout << i << ": " << lca(u, v) << " " << u << " " << v << " " << oneTop[u]
           << " " << val[u] << endl;
      assert(false);
    }
    A.push_back(an);
  }
  for (long long an : A) {
    cout << an << "\n";
  }
}