#include <bits/stdc++.h>
using namespace std;
const int N = 300100;
int tp[N], fa[N], dp[N], hc[N], sz[N], tf[N], n, q;
long long a[N], sum[N], f[N], tmp[N];
int lca(int x, int y) {
  for (; tp[x] != tp[y]; x = fa[tp[x]])
    if (dp[tp[x]] < dp[tp[y]]) swap(x, y);
  return dp[x] < dp[y] ? x : y;
}
vector<pair<int, int> > g[N];
void dfs(int x) {
  for (auto i : g[x])
    if (i.first != fa[x]) {
      fa[i.first] = x;
      dp[i.first] = dp[x] + 1;
      tf[i.first] = i.second;
      dfs(i.first);
      sz[x] += sz[i.first];
      if (sz[i.first] > sz[hc[x]]) hc[x] = i.first;
      a[x] += sum[i.first];
    }
  sz[x] += !sz[x];
  sum[x] = max(0ll, a[x] - 2 * tf[x]);
}
void dfs2(int x, int y) {
  tp[x] = y;
  f[x] = f[fa[x]] + a[fa[x]] - sum[x] - tf[x];
  tmp[x] = max(0ll, tmp[fa[x]] + a[fa[x]] - sum[x] - 2 * tf[x]);
  if (hc[x]) dfs2(hc[x], y);
  for (auto i : g[x])
    if (!tp[i.first]) dfs2(i.first, i.first);
  tmp[x] -= f[x];
  f[x] += a[x];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i < n; ++i) {
    int x, y, z;
    cin >> x >> y >> z;
    g[x].emplace_back(y, z);
    g[y].emplace_back(x, z);
  }
  dfs(1);
  dfs2(1, 1);
  for (; q-- > 0;) {
    int x, y, z;
    cin >> x >> y;
    z = lca(x, y);
    cout << (f[x] + f[y] - f[z] + tmp[z]) << '\n';
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve the following 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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 300100;
int tp[N], fa[N], dp[N], hc[N], sz[N], tf[N], n, q;
long long a[N], sum[N], f[N], tmp[N];
int lca(int x, int y) {
  for (; tp[x] != tp[y]; x = fa[tp[x]])
    if (dp[tp[x]] < dp[tp[y]]) swap(x, y);
  return dp[x] < dp[y] ? x : y;
}
vector<pair<int, int> > g[N];
void dfs(int x) {
  for (auto i : g[x])
    if (i.first != fa[x]) {
      fa[i.first] = x;
      dp[i.first] = dp[x] + 1;
      tf[i.first] = i.second;
      dfs(i.first);
      sz[x] += sz[i.first];
      if (sz[i.first] > sz[hc[x]]) hc[x] = i.first;
      a[x] += sum[i.first];
    }
  sz[x] += !sz[x];
  sum[x] = max(0ll, a[x] - 2 * tf[x]);
}
void dfs2(int x, int y) {
  tp[x] = y;
  f[x] = f[fa[x]] + a[fa[x]] - sum[x] - tf[x];
  tmp[x] = max(0ll, tmp[fa[x]] + a[fa[x]] - sum[x] - 2 * tf[x]);
  if (hc[x]) dfs2(hc[x], y);
  for (auto i : g[x])
    if (!tp[i.first]) dfs2(i.first, i.first);
  tmp[x] -= f[x];
  f[x] += a[x];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i < n; ++i) {
    int x, y, z;
    cin >> x >> y >> z;
    g[x].emplace_back(y, z);
    g[y].emplace_back(x, z);
  }
  dfs(1);
  dfs2(1, 1);
  for (; q-- > 0;) {
    int x, y, z;
    cin >> x >> y;
    z = lca(x, y);
    cout << (f[x] + f[y] - f[z] + tmp[z]) << '\n';
  }
  return 0;
}

```