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;
constexpr int N = 3e5 + 1;
constexpr int Q = 4e5 + 1;
constexpr int log2N = 20;
int n, q;
int a[N];
int depth[N];
long long gs[N];
long long ss[N];
long long toti[N];
int par[N][log2N];
vector<pair<int, int>> e[N];
vector<long long> s[N];
int qu[Q];
int qv[Q];
void par_dfs(int i) {
  for (const auto& v_w : e[i]) {
    int v = v_w.first;
    if (v == par[i][0]) {
      continue;
    }
    par[v][0] = i;
    depth[v] = depth[i] + 1;
    par_dfs(v);
  }
}
long long get_s(int i) {
  long long s_i = a[i];
  for (const auto& c : s[i]) {
    s_i += max(c, 0LL);
  }
  return s_i;
}
long long get_ans(int i) { return gs[i] + toti[i]; }
long long s_dfs(int i) {
  s[i].resize(e[i].size(), 0);
  for (size_t it = 0; it < e[i].size(); ++it) {
    const auto& v_w = e[i][it];
    int v = v_w.first;
    int w = v_w.second;
    if (v == par[i][0]) {
      continue;
    }
    s[i][it] = s_dfs(v) - 2LL * w;
  }
  return get_s(i);
}
void ss_dfs(int i, long long ans_top = 0) {
  ss[i] = max(0LL, ans_top);
  for (size_t it = 0; it < e[i].size(); ++it) {
    const auto& v_w = e[i][it];
    int v = v_w.first;
    int w = v_w.second;
    if (v == par[i][0]) {
      continue;
    }
    ss_dfs(v, ss[i] + gs[i] - 2LL * w - max(s[i][it], 0LL));
  }
}
void ans_dfs(int i, long long tot = 0LL) {
  gs[i] = get_s(i);
  toti[i] = tot;
  for (size_t it = 0; it < e[i].size(); ++it) {
    const auto& v_w = e[i][it];
    int v = v_w.first;
    int w = v_w.second;
    if (v == par[i][0]) {
      continue;
    }
    ans_dfs(v, toti[i] + gs[i] - w - max(s[i][it], 0LL));
  }
}
int lca(int u, int v) {
  if (depth[u] < depth[v]) swap(u, v);
  for (int j = log2N - 1; j >= 0; --j) {
    if (depth[par[u][j]] >= depth[v]) {
      u = par[u][j];
    }
  }
  if (u == v) return u;
  for (int j = log2N - 1; j >= 0; --j) {
    if (par[u][j] != par[v][j]) {
      u = par[u][j];
      v = par[v][j];
    }
  }
  assert(par[u][0] == par[v][0]);
  return par[u][0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    e[u].emplace_back(v, w);
    e[v].emplace_back(u, w);
  }
  for (int i = 0; i < q; ++i) {
    cin >> qu[i] >> qv[i];
  }
  depth[1] = 1;
  par_dfs(1);
  for (int j = 1; j < log2N; ++j) {
    for (int i = 1; i <= n; ++i) {
      par[i][j] = par[par[i][j - 1]][j - 1];
    }
  }
  s_dfs(1);
  ans_dfs(1);
  ss_dfs(1);
  for (int i = 0; i < q; ++i) {
    int u = qu[i];
    int v = qv[i];
    int a = lca(u, v);
    long long result = get_ans(u) + get_ans(v) - get_ans(a);
    result -= toti[a];
    result += ss[a];
    cout << result << '\n';
  }
}