#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10;
long long int dp1[N], dp2[N], Sa[N], Sw[N], dp[20][N], A[N];
int P[20][N], H[N], n, q;
vector<pair<int, int> > adj[N];
void preDFS(int v, int p = -1) {
  if (!~p) P[0][v] = v;
  for (int i = 1; i < 20; i++) P[i][v] = P[i - 1][P[i - 1][v]];
  for (pair<int, int> u : adj[v]) {
    if (u.first != p) {
      P[0][u.first] = v;
      H[u.first] = H[v] + 1;
      Sw[u.first] = Sw[v] + u.second, Sa[u.first] = Sa[v] + A[u.first];
      preDFS(u.first, v);
    }
  }
}
int getpar(int v, int h) {
  for (int i = h; i; i -= i & -i) v = P[__builtin_ctz(i)][v];
  return v;
}
int LCA(int u, int v) {
  if (H[u] < H[v]) swap(u, v);
  u = getpar(u, H[u] - H[v]);
  if (u == v) return v;
  for (int i = 19; ~i; i--)
    if (P[i][u] != P[i][v]) u = P[i][u], v = P[i][v];
  return P[0][v];
}
void DFSdn(int v, int p = -1) {
  for (pair<int, int> u : adj[v]) {
    if (u.first != p)
      DFSdn(u.first, v),
          dp1[v] += max(0ll, dp1[u.first] - 2 * u.second + A[u.first]);
  }
}
void DFSup(int v, int p = -1) {
  for (pair<int, int> u : adj[v]) {
    if (u.first != p) {
      dp2[u.first] =
          max(0ll, A[v] + dp2[v] + dp1[v] -
                       max(0ll, dp1[u.first] - 2 * u.second + A[u.first]));
      dp2[u.first] = max(0ll, dp2[u.first] - 2 * u.second);
      DFSup(u.first, v);
    }
  }
}
void NIMA(int v, int p = -1) {
  for (pair<int, int> u : adj[v]) {
    if (u.first != p) {
      dp[0][u.first] =
          dp1[v] - max(0ll, dp1[u.first] - 2 * u.second + A[u.first]);
      NIMA(u.first, v);
    }
  }
}
long long int get(int v, int h) {
  long long int ret = 0;
  for (int i = h; i; i -= i & -i)
    ret += dp[__builtin_ctz(i)][v], v = P[__builtin_ctz(i)][v];
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%lld", &A[i]);
  for (int i = 1; i < n; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    adj[u].push_back({v, w});
    adj[v].push_back({u, w});
  }
  preDFS(1);
  DFSdn(1);
  DFSup(1);
  NIMA(1);
  for (int i = 1; i < 20; i++)
    for (int j = 1; j <= n; j++)
      dp[i][j] = dp[i - 1][j] + dp[i - 1][P[i - 1][j]];
  for (; q; q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (H[u] < H[v]) swap(u, v);
    if (u == v) {
      printf("%lld\n", dp1[u] + dp2[u] + A[u]);
      continue;
    }
    int p = LCA(u, v);
    long long int a = get(u, H[u] - H[p]), b = get(v, H[v] - H[p]);
    long long int ret =
        Sa[u] - Sa[p] + A[p] + Sa[v] - Sa[p] - (Sw[u] - Sw[p] + Sw[v] - Sw[p]);
    printf("%lld\n", a + b + dp2[p] - dp1[p] + ret + dp1[u] + dp1[v]);
  }
  return 0;
}

### 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 int N = 3e5 + 10;
long long int dp1[N], dp2[N], Sa[N], Sw[N], dp[20][N], A[N];
int P[20][N], H[N], n, q;
vector<pair<int, int> > adj[N];
void preDFS(int v, int p = -1) {
  if (!~p) P[0][v] = v;
  for (int i = 1; i < 20; i++) P[i][v] = P[i - 1][P[i - 1][v]];
  for (pair<int, int> u : adj[v]) {
    if (u.first != p) {
      P[0][u.first] = v;
      H[u.first] = H[v] + 1;
      Sw[u.first] = Sw[v] + u.second, Sa[u.first] = Sa[v] + A[u.first];
      preDFS(u.first, v);
    }
  }
}
int getpar(int v, int h) {
  for (int i = h; i; i -= i & -i) v = P[__builtin_ctz(i)][v];
  return v;
}
int LCA(int u, int v) {
  if (H[u] < H[v]) swap(u, v);
  u = getpar(u, H[u] - H[v]);
  if (u == v) return v;
  for (int i = 19; ~i; i--)
    if (P[i][u] != P[i][v]) u = P[i][u], v = P[i][v];
  return P[0][v];
}
void DFSdn(int v, int p = -1) {
  for (pair<int, int> u : adj[v]) {
    if (u.first != p)
      DFSdn(u.first, v),
          dp1[v] += max(0ll, dp1[u.first] - 2 * u.second + A[u.first]);
  }
}
void DFSup(int v, int p = -1) {
  for (pair<int, int> u : adj[v]) {
    if (u.first != p) {
      dp2[u.first] =
          max(0ll, A[v] + dp2[v] + dp1[v] -
                       max(0ll, dp1[u.first] - 2 * u.second + A[u.first]));
      dp2[u.first] = max(0ll, dp2[u.first] - 2 * u.second);
      DFSup(u.first, v);
    }
  }
}
void NIMA(int v, int p = -1) {
  for (pair<int, int> u : adj[v]) {
    if (u.first != p) {
      dp[0][u.first] =
          dp1[v] - max(0ll, dp1[u.first] - 2 * u.second + A[u.first]);
      NIMA(u.first, v);
    }
  }
}
long long int get(int v, int h) {
  long long int ret = 0;
  for (int i = h; i; i -= i & -i)
    ret += dp[__builtin_ctz(i)][v], v = P[__builtin_ctz(i)][v];
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%lld", &A[i]);
  for (int i = 1; i < n; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    adj[u].push_back({v, w});
    adj[v].push_back({u, w});
  }
  preDFS(1);
  DFSdn(1);
  DFSup(1);
  NIMA(1);
  for (int i = 1; i < 20; i++)
    for (int j = 1; j <= n; j++)
      dp[i][j] = dp[i - 1][j] + dp[i - 1][P[i - 1][j]];
  for (; q; q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (H[u] < H[v]) swap(u, v);
    if (u == v) {
      printf("%lld\n", dp1[u] + dp2[u] + A[u]);
      continue;
    }
    int p = LCA(u, v);
    long long int a = get(u, H[u] - H[p]), b = get(v, H[v] - H[p]);
    long long int ret =
        Sa[u] - Sa[p] + A[p] + Sa[v] - Sa[p] - (Sw[u] - Sw[p] + Sw[v] - Sw[p]);
    printf("%lld\n", a + b + dp2[p] - dp1[p] + ret + dp1[u] + dp1[v]);
  }
  return 0;
}

```