#include <bits/stdc++.h>
using namespace std;
long long n, q;
long long a[300005], dp_child[300005], dp_par[300005];
long long now, tin[300005], tout[300005], up[300005][19], val[300005][19],
    val2[300005][19];
vector<pair<long long, long long> > g[300005];
map<pair<long long, long long>, long long> mp;
void get_child(long long u, long long p) {
  for (auto e : g[u])
    if (e.first != p) {
      long long v = e.first, w = e.second;
      get_child(v, u);
      dp_child[u] += max(0ll, dp_child[v] + a[v] - 2 * w);
      mp[{v, u}] = max(0ll, dp_child[v] + a[v] - 2 * w);
    }
}
void get_par(long long u, long long p) {
  for (auto e : g[u])
    if (e.first != p) {
      long long v = e.first, w = e.second;
      long long t = max(0ll, dp_child[u] - mp[{v, u}]);
      dp_par[v] = max(0ll, dp_par[u] + a[u] - 2 * w + t);
      get_par(v, u);
    }
}
void dfs(long long u, long long p) {
  tin[u] = ++now;
  for (long long i = 1; i <= 18; i++) {
    up[u][i] = up[up[u][i - 1]][i - 1];
    val[u][i] = val[u][i - 1] + val[up[u][i - 1]][i - 1];
    val2[u][i] = val2[u][i - 1] + val2[up[u][i - 1]][i - 1];
  }
  for (auto e : g[u])
    if (e.first != p) {
      long long v = e.first, w = e.second;
      up[v][0] = u;
      val[v][0] = a[v] - w + dp_child[v];
      val2[v][0] = mp[{v, u}];
      dfs(v, u);
    }
  tout[u] = now;
}
bool is_ancestor(long long u, long long v) {
  return (tin[u] <= tin[v] && tout[u] >= tout[v]);
}
long long lca(long long u, long long v) {
  if (is_ancestor(u, v)) return u;
  if (is_ancestor(v, u)) return v;
  for (long long i = 18; i >= 0; i--)
    if (!is_ancestor(up[u][i], v)) u = up[u][i];
  return up[u][0];
}
long long solve(long long u, long long v) {
  long long r = lca(u, v), res = 0;
  for (long long i = 18; i >= 0; i--) {
    if (!is_ancestor(up[u][i], r)) {
      res += val[u][i] - val2[u][i];
      u = up[u][i];
    }
    if (!is_ancestor(up[v][i], r)) {
      res += val[v][i] - val2[v][i];
      v = up[v][i];
    }
  }
  if (u != r) {
    res += val[u][0];
    res -= mp[{u, r}];
  }
  if (v != r) {
    res += val[v][0];
    res -= mp[{v, r}];
  }
  return res + a[r] + dp_child[r] + dp_par[r];
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n - 1; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    g[u].push_back({v, w});
    g[v].push_back({u, w});
  }
  get_child(1, 0);
  get_par(1, 0);
  tin[0] = -1000000000000000000;
  tout[0] = 1000000000000000000;
  dfs(1, 0);
  while (q--) {
    long long u, v;
    cin >> u >> v;
    cout << solve(u, v) << "\n";
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, q;
long long a[300005], dp_child[300005], dp_par[300005];
long long now, tin[300005], tout[300005], up[300005][19], val[300005][19],
    val2[300005][19];
vector<pair<long long, long long> > g[300005];
map<pair<long long, long long>, long long> mp;
void get_child(long long u, long long p) {
  for (auto e : g[u])
    if (e.first != p) {
      long long v = e.first, w = e.second;
      get_child(v, u);
      dp_child[u] += max(0ll, dp_child[v] + a[v] - 2 * w);
      mp[{v, u}] = max(0ll, dp_child[v] + a[v] - 2 * w);
    }
}
void get_par(long long u, long long p) {
  for (auto e : g[u])
    if (e.first != p) {
      long long v = e.first, w = e.second;
      long long t = max(0ll, dp_child[u] - mp[{v, u}]);
      dp_par[v] = max(0ll, dp_par[u] + a[u] - 2 * w + t);
      get_par(v, u);
    }
}
void dfs(long long u, long long p) {
  tin[u] = ++now;
  for (long long i = 1; i <= 18; i++) {
    up[u][i] = up[up[u][i - 1]][i - 1];
    val[u][i] = val[u][i - 1] + val[up[u][i - 1]][i - 1];
    val2[u][i] = val2[u][i - 1] + val2[up[u][i - 1]][i - 1];
  }
  for (auto e : g[u])
    if (e.first != p) {
      long long v = e.first, w = e.second;
      up[v][0] = u;
      val[v][0] = a[v] - w + dp_child[v];
      val2[v][0] = mp[{v, u}];
      dfs(v, u);
    }
  tout[u] = now;
}
bool is_ancestor(long long u, long long v) {
  return (tin[u] <= tin[v] && tout[u] >= tout[v]);
}
long long lca(long long u, long long v) {
  if (is_ancestor(u, v)) return u;
  if (is_ancestor(v, u)) return v;
  for (long long i = 18; i >= 0; i--)
    if (!is_ancestor(up[u][i], v)) u = up[u][i];
  return up[u][0];
}
long long solve(long long u, long long v) {
  long long r = lca(u, v), res = 0;
  for (long long i = 18; i >= 0; i--) {
    if (!is_ancestor(up[u][i], r)) {
      res += val[u][i] - val2[u][i];
      u = up[u][i];
    }
    if (!is_ancestor(up[v][i], r)) {
      res += val[v][i] - val2[v][i];
      v = up[v][i];
    }
  }
  if (u != r) {
    res += val[u][0];
    res -= mp[{u, r}];
  }
  if (v != r) {
    res += val[v][0];
    res -= mp[{v, r}];
  }
  return res + a[r] + dp_child[r] + dp_par[r];
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n - 1; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    g[u].push_back({v, w});
    g[v].push_back({u, w});
  }
  get_child(1, 0);
  get_par(1, 0);
  tin[0] = -1000000000000000000;
  tout[0] = 1000000000000000000;
  dfs(1, 0);
  while (q--) {
    long long u, v;
    cin >> u >> v;
    cout << solve(u, v) << "\n";
  }
}

```