#include <bits/stdc++.h>
using namespace std;
const int N = 400005;
int n, q;
vector<int> a(N);
vector<pair<int, int>> edges[N];
vector<vector<int>> prt(N, vector<int>(20));
vector<long long> dp1(N, 0);
vector<long long> dp2(N, 0);
vector<vector<long long>> dp3(N);
vector<long long> dp4(N, 0);
vector<int> tin(N, 0);
vector<int> tout(N, 0);
int T = 1;
void dfs(int s, int p) {
  tin[s] = T++;
  prt[s][0] = p;
  for (int st = 1; st < 20; st += 1) {
    prt[s][st] = prt[prt[s][st - 1]][st - 1];
  }
  dp1[s] = a[s];
  for (auto &next : edges[s]) {
    int v = next.first;
    int w = next.second;
    if (v == p) continue;
    dfs(v, s);
    dp1[s] += max(0ll, dp1[v] - 2ll * w);
  }
  tout[s] = T++;
}
void cal(int s, int p, long long rtt) {
  dp2[s] = a[s];
  dp4[s] = rtt + dp1[s];
  for (auto &next : edges[s]) {
    int v = next.first;
    int w = next.second;
    long long tmp = max(0ll, dp1[v] - 2ll * w);
    dp2[s] += tmp;
    dp3[s].push_back(tmp);
  }
  int i = -1;
  for (auto &next : edges[s]) {
    i += 1;
    int v = next.first;
    int w = next.second;
    if (v == p) continue;
    long long tmp = dp1[s];
    dp1[s] = dp2[s] - dp3[s][i];
    cal(v, s, dp4[s] - dp3[s][i] - w);
    dp1[s] = tmp;
  }
}
bool isP(int u, int v) { return tin[u] <= tin[v] && tout[v] <= tout[u]; }
int lca(int u, int v) {
  if (isP(u, v)) return u;
  if (isP(v, u)) return v;
  for (int st = 19; st >= 0; st -= 1) {
    if (!isP(prt[u][st], v)) u = prt[u][st];
  }
  return prt[u][0];
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i += 1) scanf("%d", &a[i]);
  for (int i = 1; i < n; i += 1) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    edges[u].emplace_back(v, w);
    edges[v].emplace_back(u, w);
  }
  dfs(1, 1);
  cal(1, 1, 0);
  for (int i = 0; i < q; i += 1) {
    int u, v;
    scanf("%d %d", &u, &v);
    int cp = lca(u, v);
    long long ans = dp2[cp] + (dp4[u] - dp4[cp]) + (dp4[v] - dp4[cp]);
    printf("%I64d\n", ans);
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 = 400005;
int n, q;
vector<int> a(N);
vector<pair<int, int>> edges[N];
vector<vector<int>> prt(N, vector<int>(20));
vector<long long> dp1(N, 0);
vector<long long> dp2(N, 0);
vector<vector<long long>> dp3(N);
vector<long long> dp4(N, 0);
vector<int> tin(N, 0);
vector<int> tout(N, 0);
int T = 1;
void dfs(int s, int p) {
  tin[s] = T++;
  prt[s][0] = p;
  for (int st = 1; st < 20; st += 1) {
    prt[s][st] = prt[prt[s][st - 1]][st - 1];
  }
  dp1[s] = a[s];
  for (auto &next : edges[s]) {
    int v = next.first;
    int w = next.second;
    if (v == p) continue;
    dfs(v, s);
    dp1[s] += max(0ll, dp1[v] - 2ll * w);
  }
  tout[s] = T++;
}
void cal(int s, int p, long long rtt) {
  dp2[s] = a[s];
  dp4[s] = rtt + dp1[s];
  for (auto &next : edges[s]) {
    int v = next.first;
    int w = next.second;
    long long tmp = max(0ll, dp1[v] - 2ll * w);
    dp2[s] += tmp;
    dp3[s].push_back(tmp);
  }
  int i = -1;
  for (auto &next : edges[s]) {
    i += 1;
    int v = next.first;
    int w = next.second;
    if (v == p) continue;
    long long tmp = dp1[s];
    dp1[s] = dp2[s] - dp3[s][i];
    cal(v, s, dp4[s] - dp3[s][i] - w);
    dp1[s] = tmp;
  }
}
bool isP(int u, int v) { return tin[u] <= tin[v] && tout[v] <= tout[u]; }
int lca(int u, int v) {
  if (isP(u, v)) return u;
  if (isP(v, u)) return v;
  for (int st = 19; st >= 0; st -= 1) {
    if (!isP(prt[u][st], v)) u = prt[u][st];
  }
  return prt[u][0];
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i += 1) scanf("%d", &a[i]);
  for (int i = 1; i < n; i += 1) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    edges[u].emplace_back(v, w);
    edges[v].emplace_back(u, w);
  }
  dfs(1, 1);
  cal(1, 1, 0);
  for (int i = 0; i < q; i += 1) {
    int u, v;
    scanf("%d %d", &u, &v);
    int cp = lca(u, v);
    long long ans = dp2[cp] + (dp4[u] - dp4[cp]) + (dp4[v] - dp4[cp]);
    printf("%I64d\n", ans);
  }
}

```