#include <bits/stdc++.h>
using namespace std;
const int MaxN = 3e5 + 6.66;
const int MaxLog = 20;
int up_ed[MaxN];
long long dp[MaxN];
long long dp_up[MaxN];
long long ret_dp[MaxN];
long long lca_sum[MaxLog][MaxN];
long long lca_rem[MaxLog][MaxN];
int lca[MaxLog][MaxN];
int arr[MaxN];
int h[MaxN];
vector<pair<int, int>> tree[MaxN];
long long dfs_up(int v, int par = 0, int up = 0, int hei = 0) {
  lca[0][v] = par;
  up_ed[v] = up;
  h[v] = hei;
  dp[v] = arr[v];
  for (auto x : tree[v])
    if (x.first != par) dp[v] += dfs_up(x.first, v, x.second, hei + 1);
  return ret_dp[v] = max(0LL, dp[v] - 2 * up);
}
void dfs_down(int v, int par = 0, long long up_val = 0) {
  dp_up[v] = up_val;
  long long suf_sum = arr[v] + up_val;
  for (auto u : tree[v])
    if (u.first != par) suf_sum += ret_dp[u.first];
  long long pre_sum = 0;
  for (auto u : tree[v]) {
    if (u.first == par) continue;
    suf_sum -= ret_dp[u.first];
    dfs_down(u.first, v, max(0LL, pre_sum + suf_sum - 2 * u.second));
    pre_sum += ret_dp[u.first];
  }
}
void init_lca(int n) {
  for (int i = 1; i <= n; i++)
    lca_sum[0][i] = dp[i] - up_ed[i], lca_rem[0][i] = ret_dp[i];
  for (int i = 1; i < MaxLog; i++)
    for (int j = 1; j <= n; j++) {
      lca[i][j] = lca[i - 1][lca[i - 1][j]];
      lca_sum[i][j] = lca_sum[i - 1][j] + lca_sum[i - 1][lca[i - 1][j]];
      lca_rem[i][j] = lca_rem[i - 1][j] + lca_rem[i - 1][lca[i - 1][j]];
    }
}
int get_par(int v, int x) {
  for (int i = 0; i < MaxLog; i++)
    if (x & (1 << i)) v = lca[i][v];
  return v;
}
long long get(int u, int v) {
  long long ans = 0;
  if (h[v] > h[u]) swap(u, v);
  for (int i = MaxLog - 1; i > -1; --i)
    if (h[u] - h[v] >= (1 << i)) {
      ans += lca_sum[i][u] - lca_rem[i][u];
      u = lca[i][u];
    }
  if (u == v) return ans + dp[u] + dp_up[u];
  int v0 = v;
  for (int i = MaxLog - 1; i > -1; --i)
    if ((1 << i) <= h[v] && lca[i][v] != lca[i][u]) {
      ans += lca_sum[i][u] + lca_sum[i][v] - lca_rem[i][u] - lca_rem[i][v];
      u = lca[i][u];
      v = lca[i][v];
    }
  ans += lca_sum[0][u] + lca_sum[0][v] - lca_rem[0][u] - lca_rem[0][v];
  int l = lca[0][v];
  return ans + dp[l] + dp_up[l];
}
void init(int n) {
  dfs_up(1);
  dfs_down(1);
  init_lca(n);
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i);
  for (int rki = 1; rki < n; rki++) {
    int i, j, v;
    scanf("%d %d %d", &i, &j, &v);
    tree[i].push_back({j, v});
    tree[j].push_back({i, v});
  }
  init(n);
  while (q--) {
    int u, v;
    scanf("%d %d", &u, &v);
    printf("%lld\n", get(u, v));
  }
  return 0;
}

### 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 MaxN = 3e5 + 6.66;
const int MaxLog = 20;
int up_ed[MaxN];
long long dp[MaxN];
long long dp_up[MaxN];
long long ret_dp[MaxN];
long long lca_sum[MaxLog][MaxN];
long long lca_rem[MaxLog][MaxN];
int lca[MaxLog][MaxN];
int arr[MaxN];
int h[MaxN];
vector<pair<int, int>> tree[MaxN];
long long dfs_up(int v, int par = 0, int up = 0, int hei = 0) {
  lca[0][v] = par;
  up_ed[v] = up;
  h[v] = hei;
  dp[v] = arr[v];
  for (auto x : tree[v])
    if (x.first != par) dp[v] += dfs_up(x.first, v, x.second, hei + 1);
  return ret_dp[v] = max(0LL, dp[v] - 2 * up);
}
void dfs_down(int v, int par = 0, long long up_val = 0) {
  dp_up[v] = up_val;
  long long suf_sum = arr[v] + up_val;
  for (auto u : tree[v])
    if (u.first != par) suf_sum += ret_dp[u.first];
  long long pre_sum = 0;
  for (auto u : tree[v]) {
    if (u.first == par) continue;
    suf_sum -= ret_dp[u.first];
    dfs_down(u.first, v, max(0LL, pre_sum + suf_sum - 2 * u.second));
    pre_sum += ret_dp[u.first];
  }
}
void init_lca(int n) {
  for (int i = 1; i <= n; i++)
    lca_sum[0][i] = dp[i] - up_ed[i], lca_rem[0][i] = ret_dp[i];
  for (int i = 1; i < MaxLog; i++)
    for (int j = 1; j <= n; j++) {
      lca[i][j] = lca[i - 1][lca[i - 1][j]];
      lca_sum[i][j] = lca_sum[i - 1][j] + lca_sum[i - 1][lca[i - 1][j]];
      lca_rem[i][j] = lca_rem[i - 1][j] + lca_rem[i - 1][lca[i - 1][j]];
    }
}
int get_par(int v, int x) {
  for (int i = 0; i < MaxLog; i++)
    if (x & (1 << i)) v = lca[i][v];
  return v;
}
long long get(int u, int v) {
  long long ans = 0;
  if (h[v] > h[u]) swap(u, v);
  for (int i = MaxLog - 1; i > -1; --i)
    if (h[u] - h[v] >= (1 << i)) {
      ans += lca_sum[i][u] - lca_rem[i][u];
      u = lca[i][u];
    }
  if (u == v) return ans + dp[u] + dp_up[u];
  int v0 = v;
  for (int i = MaxLog - 1; i > -1; --i)
    if ((1 << i) <= h[v] && lca[i][v] != lca[i][u]) {
      ans += lca_sum[i][u] + lca_sum[i][v] - lca_rem[i][u] - lca_rem[i][v];
      u = lca[i][u];
      v = lca[i][v];
    }
  ans += lca_sum[0][u] + lca_sum[0][v] - lca_rem[0][u] - lca_rem[0][v];
  int l = lca[0][v];
  return ans + dp[l] + dp_up[l];
}
void init(int n) {
  dfs_up(1);
  dfs_down(1);
  init_lca(n);
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i);
  for (int rki = 1; rki < n; rki++) {
    int i, j, v;
    scanf("%d %d %d", &i, &j, &v);
    tree[i].push_back({j, v});
    tree[j].push_back({i, v});
  }
  init(n);
  while (q--) {
    int u, v;
    scanf("%d %d", &u, &v);
    printf("%lld\n", get(u, v));
  }
  return 0;
}

```