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;
using int64 = long long int;
struct Edge {
  int v;
  int w;
};
class Tree {
 public:
  Tree(const vector<int>& a)
      : a_(a),
        edges_(a_.size()),
        enter_time_(a_.size()),
        leave_time_(a_.size()),
        f1_(a_.size()),
        f2_(a_.size()),
        path_sum_(a_.size()),
        outside_(a_.size()) {}
  void AddEdge(int u, int v, int w) {
    edges_[u].push_back({v, w});
    edges_[v].push_back({u, w});
  }
  void Build() {
    Dfs(0, -1);
    BuildLcaTree();
    outside_[0] = 0;
    DfsOutside(0, -1, 0);
  }
  int64 Query(int u, int v) {
    const int l = lca(u, v);
    return path_sum_[u] + path_sum_[v] - 2 * path_sum_[l] + f1_[l] +
           outside_[l];
  }
  int lca(int u, int v) {
    if (enter_time_[u] > enter_time_[v]) {
      swap(u, v);
    }
    u = lca_offset_ + enter_time_[u];
    v = lca_offset_ + enter_time_[v];
    int r = lca_tree_[u];
    while (u <= v) {
      if (u % 2 == 0) {
        if (enter_time_[r] > enter_time_[lca_tree_[u]]) {
          r = lca_tree_[u];
        }
        ++u;
      }
      if (v % 2 == 1) {
        if (enter_time_[r] > enter_time_[lca_tree_[v]]) {
          r = lca_tree_[v];
        }
        --v;
      }
      u = (u - 1) >> 1;
      v = (v - 1) >> 1;
    }
    return r;
  }

 private:
  void Dfs(int u, int p) {
    enter_time_[u] = lca_sequence_.size();
    lca_sequence_.push_back(u);
    f1_[u] = a_[u];
    for (const Edge& e : edges_[u]) {
      if (e.v != p) {
        Dfs(e.v, u);
        lca_sequence_.push_back(u);
        f2_[e.v] = max(0LL, f1_[e.v] - 2 * e.w);
        f1_[u] += f2_[e.v];
      }
    }
    leave_time_[u] = lca_sequence_.size();
  }
  void BuildLcaTree() {
    lca_offset_ = 1;
    while (lca_offset_ < lca_sequence_.size()) {
      lca_offset_ <<= 1;
    }
    while (lca_sequence_.size() < lca_offset_) {
      lca_sequence_.push_back(lca_sequence_.back());
    }
    --lca_offset_;
    lca_tree_.assign(2 * lca_offset_ + 1, -1);
    for (size_t i = 0; i <= lca_offset_; ++i) {
      lca_tree_[lca_offset_ + i] = lca_sequence_[i];
    }
    for (int i = lca_offset_ - 1; i >= 0; --i) {
      int left = (i << 1) + 1;
      int right = left + 1;
      if (enter_time_[lca_tree_[left]] < enter_time_[lca_tree_[right]]) {
        lca_tree_[i] = lca_tree_[left];
      } else {
        lca_tree_[i] = lca_tree_[right];
      }
    }
  }
  void DfsOutside(int u, int p, int64 sum) {
    path_sum_[u] = f1_[u] - f2_[u] + sum;
    for (const Edge& e : edges_[u]) {
      if (e.v != p) {
        outside_[e.v] = max(0LL, -2 * e.w + f1_[u] - f2_[e.v] + outside_[u]);
        DfsOutside(e.v, u, path_sum_[u] - e.w);
      }
    }
  }
  const vector<int> a_;
  vector<vector<Edge>> edges_;
  vector<int> enter_time_;
  vector<int> leave_time_;
  vector<int> lca_sequence_;
  vector<int> lca_tree_;
  size_t lca_offset_;
  vector<int64> f1_;
  vector<int64> f2_;
  vector<int64> path_sum_;
  vector<int64> outside_;
};
void solve() {
  int n, q;
  scanf("%d %d", &n, &q);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  Tree t(a);
  for (int i = 0; i < n - 1; ++i) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    t.AddEdge(u - 1, v - 1, w);
  }
  t.Build();
  for (int i = 0; i < q; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    cout << t.Query(u - 1, v - 1) << "\n";
  }
}
int main() {
  solve();
  return 0;
}