#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 10, M = 3e5 + 100, MOD = 1e9 + 7, ML = 25;
long long a[M], dp[M], dpu[M], is[M], h[M], lca[ML][M], lc[ML][M];
vector<pair<long long, long long> > adj[M];
void dfs(int v, int p) {
  dp[v] = a[v];
  h[v] = h[p] + 1;
  for (pair<long long, long long> u : adj[v]) {
    if (u.first == p) continue;
    dfs(u.first, v);
    if (dp[u.first] - u.second * 2 >= 0) dp[v] += dp[u.first] - u.second * 2;
  }
}
void dfsu(int v, int p, long long w) {
  if (p != v) {
    long long x = dpu[p] + dp[p];
    if (w * 2 < dp[v]) {
      x -= dp[v] - w * 2;
      is[v] = 1;
    }
    if (w * 2 < x) dpu[v] = x - w * 2;
  }
  lca[0][v] = p;
  lc[0][v] = dp[v] - w;
  if (is[v]) lc[0][v] -= dp[v] - w * 2;
  for (int i = 1; i <= 20; i++) {
    lca[i][v] = lca[i - 1][lca[i - 1][v]];
    lc[i][v] = lc[i - 1][v] + lc[i - 1][lca[i - 1][v]];
  }
  for (pair<long long, long long> u : adj[v]) {
    if (u.first != p) dfsu(u.first, v, u.second);
  }
}
pair<long long, long long> gp(int v, int x) {
  long long z = 0;
  for (int i = 0; i <= 20; i++) {
    if (x & (1 << i)) {
      z += lc[i][v];
      v = lca[i][v];
    }
  }
  return {v, z};
}
long long g(int v, int u) {
  if (h[u] < h[v]) swap(v, u);
  pair<long long, long long> p = gp(u, h[u] - h[v]);
  long long ans = p.second;
  u = p.first;
  if (u == v) return ans + dp[v] + dpu[v];
  for (int i = 20; i >= 0; i--) {
    if (lca[i][v] != lca[i][u]) {
      ans += lc[i][v] + lc[i][u];
      v = lca[i][v], u = lca[i][u];
    }
  }
  ans += dp[lca[0][v]] + dpu[lca[0][v]] + lc[0][v] + lc[0][u];
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    long long x, y, z;
    cin >> x >> y >> z;
    adj[x].push_back({y, z});
    adj[y].push_back({x, z});
  }
  dfs(1, 1);
  dfsu(1, 1, 0);
  while (m--) {
    long long x, y;
    cin >> x >> y;
    cout << g(x, y) << "\n";
  }
}

### Prompt

Please formulate a CPP solution to 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 long long INF = 1e9 + 10, M = 3e5 + 100, MOD = 1e9 + 7, ML = 25;
long long a[M], dp[M], dpu[M], is[M], h[M], lca[ML][M], lc[ML][M];
vector<pair<long long, long long> > adj[M];
void dfs(int v, int p) {
  dp[v] = a[v];
  h[v] = h[p] + 1;
  for (pair<long long, long long> u : adj[v]) {
    if (u.first == p) continue;
    dfs(u.first, v);
    if (dp[u.first] - u.second * 2 >= 0) dp[v] += dp[u.first] - u.second * 2;
  }
}
void dfsu(int v, int p, long long w) {
  if (p != v) {
    long long x = dpu[p] + dp[p];
    if (w * 2 < dp[v]) {
      x -= dp[v] - w * 2;
      is[v] = 1;
    }
    if (w * 2 < x) dpu[v] = x - w * 2;
  }
  lca[0][v] = p;
  lc[0][v] = dp[v] - w;
  if (is[v]) lc[0][v] -= dp[v] - w * 2;
  for (int i = 1; i <= 20; i++) {
    lca[i][v] = lca[i - 1][lca[i - 1][v]];
    lc[i][v] = lc[i - 1][v] + lc[i - 1][lca[i - 1][v]];
  }
  for (pair<long long, long long> u : adj[v]) {
    if (u.first != p) dfsu(u.first, v, u.second);
  }
}
pair<long long, long long> gp(int v, int x) {
  long long z = 0;
  for (int i = 0; i <= 20; i++) {
    if (x & (1 << i)) {
      z += lc[i][v];
      v = lca[i][v];
    }
  }
  return {v, z};
}
long long g(int v, int u) {
  if (h[u] < h[v]) swap(v, u);
  pair<long long, long long> p = gp(u, h[u] - h[v]);
  long long ans = p.second;
  u = p.first;
  if (u == v) return ans + dp[v] + dpu[v];
  for (int i = 20; i >= 0; i--) {
    if (lca[i][v] != lca[i][u]) {
      ans += lc[i][v] + lc[i][u];
      v = lca[i][v], u = lca[i][u];
    }
  }
  ans += dp[lca[0][v]] + dpu[lca[0][v]] + lc[0][v] + lc[0][u];
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    long long x, y, z;
    cin >> x >> y >> z;
    adj[x].push_back({y, z});
    adj[y].push_back({x, z});
  }
  dfs(1, 1);
  dfsu(1, 1, 0);
  while (m--) {
    long long x, y;
    cin >> x >> y;
    cout << g(x, y) << "\n";
  }
}

```