#include <bits/stdc++.h>
using namespace std;
const long long N = 3e5 + 5;
vector<pair<long long, long long>> adj[N];
long long val[N], dp[N], head[N], sz[N], dp2[N], heavy[N], dep[N], sum[N],
    fa[N], to[N];
void dfs1(long long u, long long p) {
  long long mx = 0;
  dp[u] = val[u], sz[u] = 1;
  for (auto [v, w] : adj[u]) {
    if (v == p) continue;
    dep[v] = dep[u] + 1;
    fa[v] = u;
    to[v] = w;
    dfs1(v, u);
    sz[u] += sz[v];
    dp[u] += max((long long)0, dp[v] - w * 2);
    if (sz[v] > mx) heavy[u] = v, mx = sz[v];
  }
}
void dfs2(long long u, long long h) {
  head[u] = h;
  sum[u] =
      sum[fa[u]] + dp[fa[u]] - max((long long)0, dp[u] - to[u] * 2) - to[u];
  dp2[u] = max((long long)0, dp2[fa[u]] + dp[fa[u]] -
                                 max((long long)0, dp[u] - to[u] * 2) - to[u] -
                                 to[u]);
  if (heavy[u] != 0) dfs2(heavy[u], h);
  for (auto [v, w] : adj[u]) {
    if (v == fa[u] || v == heavy[u]) continue;
    dfs2(v, v);
  }
}
long long LCA(long long u, long long v) {
  while (head[u] != head[v]) {
    dep[head[u]] > dep[head[v]] ? u = fa[head[u]] : v = fa[head[v]];
  }
  return dep[u] < dep[v] ? u : v;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, q;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> val[i];
  for (long long i = 2; i <= n; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    adj[u].push_back({v, w});
    adj[v].push_back({u, w});
  }
  dfs1(1, 1);
  dfs2(1, 1);
  cout << "\n";
  while (q--) {
    long long u, v;
    cin >> u >> v;
    long long l = LCA(u, v);
    cout << (sum[u] - sum[l]) + (sum[v] - sum[l]) + dp[u] + dp[v] - dp[l] +
                dp2[l]
         << "\n";
  }
}

### Prompt

Your challenge is to write 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 N = 3e5 + 5;
vector<pair<long long, long long>> adj[N];
long long val[N], dp[N], head[N], sz[N], dp2[N], heavy[N], dep[N], sum[N],
    fa[N], to[N];
void dfs1(long long u, long long p) {
  long long mx = 0;
  dp[u] = val[u], sz[u] = 1;
  for (auto [v, w] : adj[u]) {
    if (v == p) continue;
    dep[v] = dep[u] + 1;
    fa[v] = u;
    to[v] = w;
    dfs1(v, u);
    sz[u] += sz[v];
    dp[u] += max((long long)0, dp[v] - w * 2);
    if (sz[v] > mx) heavy[u] = v, mx = sz[v];
  }
}
void dfs2(long long u, long long h) {
  head[u] = h;
  sum[u] =
      sum[fa[u]] + dp[fa[u]] - max((long long)0, dp[u] - to[u] * 2) - to[u];
  dp2[u] = max((long long)0, dp2[fa[u]] + dp[fa[u]] -
                                 max((long long)0, dp[u] - to[u] * 2) - to[u] -
                                 to[u]);
  if (heavy[u] != 0) dfs2(heavy[u], h);
  for (auto [v, w] : adj[u]) {
    if (v == fa[u] || v == heavy[u]) continue;
    dfs2(v, v);
  }
}
long long LCA(long long u, long long v) {
  while (head[u] != head[v]) {
    dep[head[u]] > dep[head[v]] ? u = fa[head[u]] : v = fa[head[v]];
  }
  return dep[u] < dep[v] ? u : v;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, q;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> val[i];
  for (long long i = 2; i <= n; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    adj[u].push_back({v, w});
    adj[v].push_back({u, w});
  }
  dfs1(1, 1);
  dfs2(1, 1);
  cout << "\n";
  while (q--) {
    long long u, v;
    cin >> u >> v;
    long long l = LCA(u, v);
    cout << (sum[u] - sum[l]) + (sum[v] - sum[l]) + dp[u] + dp[v] - dp[l] +
                dp2[l]
         << "\n";
  }
}

```