#include <bits/stdc++.h>
using namespace std;
const int N = 300010;
int lca[N][20];
long long sums[N][20];
long long dp[N], fa[N];
int dep[N], a[N];
vector<pair<int, int> > v[N];
void dfs1(int u, int fa) {
  dp[u] = a[u];
  for (int i = 0; i < v[u].size(); i++) {
    int first = v[u][i].first, second = v[u][i].second;
    if (first == fa) continue;
    dep[first] = dep[u] + 1;
    lca[first][0] = u;
    dfs1(first, u);
    if (dp[first] - 2 * second > 0) dp[u] += dp[first] - 2 * second;
  }
}
void dfs2(int u, int pre) {
  for (int i = 0; i < v[u].size(); i++) {
    int first = v[u][i].first, second = v[u][i].second;
    if (first == pre) continue;
    long long p = max(0LL, dp[first] - 2 * second);
    fa[first] = max(0LL, fa[u] + dp[u] - p - 2 * second);
    sums[first][0] = dp[u] - p - second;
    dfs2(first, u);
  }
}
void init_lca(int n) {
  for (int i = 1; i < 20; i++) {
    for (int j = 1; j <= n; j++) {
      lca[j][i] = lca[lca[j][i - 1]][i - 1];
      sums[j][i] = sums[j][i - 1] + sums[lca[j][i - 1]][i - 1];
    }
  }
}
long long get_ans(int first, int second) {
  if (dep[first] < dep[second]) swap(first, second);
  if (first == second) return dp[first] + fa[first];
  long long ans = dp[first];
  for (int i = 19; i >= 0; i--) {
    if (dep[first] - (1 << i) >= dep[second]) {
      ans += sums[first][i];
      first = lca[first][i];
    }
  }
  if (first == second) return ans + fa[second];
  ans += dp[second];
  for (int i = 19; i >= 0; i--) {
    if (lca[first][i] == lca[second][i]) continue;
    ans += sums[first][i] + sums[second][i];
    first = lca[first][i];
    second = lca[second][i];
  }
  int z = lca[first][0];
  ans += fa[z] - dp[z] + sums[first][0] + sums[second][0];
  return ans;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    int first, second, w;
    scanf("%d%d%d", &first, &second, &w);
    v[first].push_back(make_pair(second, w));
    v[second].push_back(make_pair(first, w));
  }
  dfs1(1, 0);
  dfs2(1, 0);
  init_lca(n);
  while (q--) {
    int first, second;
    scanf("%d%d", &first, &second);
    long long ans = get_ans(first, second);
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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 = 300010;
int lca[N][20];
long long sums[N][20];
long long dp[N], fa[N];
int dep[N], a[N];
vector<pair<int, int> > v[N];
void dfs1(int u, int fa) {
  dp[u] = a[u];
  for (int i = 0; i < v[u].size(); i++) {
    int first = v[u][i].first, second = v[u][i].second;
    if (first == fa) continue;
    dep[first] = dep[u] + 1;
    lca[first][0] = u;
    dfs1(first, u);
    if (dp[first] - 2 * second > 0) dp[u] += dp[first] - 2 * second;
  }
}
void dfs2(int u, int pre) {
  for (int i = 0; i < v[u].size(); i++) {
    int first = v[u][i].first, second = v[u][i].second;
    if (first == pre) continue;
    long long p = max(0LL, dp[first] - 2 * second);
    fa[first] = max(0LL, fa[u] + dp[u] - p - 2 * second);
    sums[first][0] = dp[u] - p - second;
    dfs2(first, u);
  }
}
void init_lca(int n) {
  for (int i = 1; i < 20; i++) {
    for (int j = 1; j <= n; j++) {
      lca[j][i] = lca[lca[j][i - 1]][i - 1];
      sums[j][i] = sums[j][i - 1] + sums[lca[j][i - 1]][i - 1];
    }
  }
}
long long get_ans(int first, int second) {
  if (dep[first] < dep[second]) swap(first, second);
  if (first == second) return dp[first] + fa[first];
  long long ans = dp[first];
  for (int i = 19; i >= 0; i--) {
    if (dep[first] - (1 << i) >= dep[second]) {
      ans += sums[first][i];
      first = lca[first][i];
    }
  }
  if (first == second) return ans + fa[second];
  ans += dp[second];
  for (int i = 19; i >= 0; i--) {
    if (lca[first][i] == lca[second][i]) continue;
    ans += sums[first][i] + sums[second][i];
    first = lca[first][i];
    second = lca[second][i];
  }
  int z = lca[first][0];
  ans += fa[z] - dp[z] + sums[first][0] + sums[second][0];
  return ans;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    int first, second, w;
    scanf("%d%d%d", &first, &second, &w);
    v[first].push_back(make_pair(second, w));
    v[second].push_back(make_pair(first, w));
  }
  dfs1(1, 0);
  dfs2(1, 0);
  init_lca(n);
  while (q--) {
    int first, second;
    scanf("%d%d", &first, &second);
    long long ans = get_ans(first, second);
    printf("%I64d\n", ans);
  }
  return 0;
}

```