#include <bits/stdc++.h>
using namespace std;
int n, q;
int value[300007];
vector<pair<int, int> > v[300007];
long long dp[300007];
long long up[300007];
int lvl[300007];
int LCA[300007][21];
long long prefix[300007];
void dfs_bot(int vertex, int prv) {
  int i;
  int sz = v[vertex].size();
  dp[vertex] = value[vertex];
  if (prv != -1) {
    for (i = 1; i < 21; i++) {
      LCA[vertex][i] = LCA[LCA[vertex][i - 1]][i - 1];
    }
  }
  for (i = 0; i < sz; i++) {
    if (v[vertex][i].first == prv) {
      continue;
    }
    lvl[v[vertex][i].first] = lvl[vertex] + 1;
    LCA[v[vertex][i].first][0] = vertex;
    dfs_bot(v[vertex][i].first, vertex);
    long long aux = dp[v[vertex][i].first] - 2 * v[vertex][i].second;
    if (aux < 0) {
      aux = 0;
    }
    dp[vertex] += aux;
  }
}
void dfs_top(int vertex, int prv, long long prv_edge) {
  up[vertex] = dp[vertex];
  if (prv != -1) {
    long long aux = up[vertex] + up[prv];
    aux -= 2 * prv_edge;
    aux -= max(dp[vertex] - 2 * prv_edge, 0LL);
    if (up[vertex] < aux) {
      up[vertex] = aux;
    }
  }
  int i;
  int sz = v[vertex].size();
  for (i = 0; i < sz; i++) {
    if (v[vertex][i].first == prv) {
      continue;
    }
    dfs_top(v[vertex][i].first, vertex, v[vertex][i].second);
  }
}
int get_lca(int x, int y) {
  int i;
  for (i = 21 - 1; i >= 0; i--) {
    if (lvl[x] - (1 << i) >= lvl[y]) {
      x = LCA[x][i];
    }
    if (lvl[y] - (1 << i) >= lvl[x]) {
      y = LCA[y][i];
    }
  }
  for (i = 21 - 1; i >= 0; i--) {
    if (LCA[x][i] != LCA[y][i]) {
      x = LCA[x][i];
      y = LCA[y][i];
    }
  }
  if (x != y) {
    x = LCA[x][0];
  }
  return x;
}
void calculate_prefixes(int vertex, int prv, int prv_edge) {
  int i;
  int sz = v[vertex].size();
  if (prv == -1) {
    prefix[vertex] = 0;
  } else {
    prefix[vertex] = prefix[prv] - prv_edge;
  }
  prefix[vertex] += dp[vertex];
  if (prv != -1) {
    long long aux = dp[vertex] - 2 * prv_edge;
    if (aux < 0) {
      aux = 0;
    }
    prefix[vertex] -= aux;
  }
  for (i = 0; i < sz; i++) {
    if (v[vertex][i].first == prv) {
      continue;
    }
    calculate_prefixes(v[vertex][i].first, vertex, v[vertex][i].second);
  }
}
void input() {
  scanf("%d%d", &n, &q);
  int i;
  for (i = 1; i <= n; i++) {
    scanf("%d", &value[i]);
  }
  for (i = 1; i < n; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    v[x].push_back(make_pair(y, z));
    v[y].push_back(make_pair(x, z));
  }
}
void solve() {
  dfs_bot(1, -1);
  dfs_top(1, -1, 0);
  calculate_prefixes(1, -1, 0);
  while (q != 0) {
    q--;
    int x, y;
    scanf("%d%d", &x, &y);
    int u = get_lca(x, y);
    long long ans = up[u];
    ans += prefix[x] - prefix[u];
    ans += prefix[y] - prefix[u];
    printf("%I64d\n", ans);
  }
}
int main() {
  input();
  solve();
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int n, q;
int value[300007];
vector<pair<int, int> > v[300007];
long long dp[300007];
long long up[300007];
int lvl[300007];
int LCA[300007][21];
long long prefix[300007];
void dfs_bot(int vertex, int prv) {
  int i;
  int sz = v[vertex].size();
  dp[vertex] = value[vertex];
  if (prv != -1) {
    for (i = 1; i < 21; i++) {
      LCA[vertex][i] = LCA[LCA[vertex][i - 1]][i - 1];
    }
  }
  for (i = 0; i < sz; i++) {
    if (v[vertex][i].first == prv) {
      continue;
    }
    lvl[v[vertex][i].first] = lvl[vertex] + 1;
    LCA[v[vertex][i].first][0] = vertex;
    dfs_bot(v[vertex][i].first, vertex);
    long long aux = dp[v[vertex][i].first] - 2 * v[vertex][i].second;
    if (aux < 0) {
      aux = 0;
    }
    dp[vertex] += aux;
  }
}
void dfs_top(int vertex, int prv, long long prv_edge) {
  up[vertex] = dp[vertex];
  if (prv != -1) {
    long long aux = up[vertex] + up[prv];
    aux -= 2 * prv_edge;
    aux -= max(dp[vertex] - 2 * prv_edge, 0LL);
    if (up[vertex] < aux) {
      up[vertex] = aux;
    }
  }
  int i;
  int sz = v[vertex].size();
  for (i = 0; i < sz; i++) {
    if (v[vertex][i].first == prv) {
      continue;
    }
    dfs_top(v[vertex][i].first, vertex, v[vertex][i].second);
  }
}
int get_lca(int x, int y) {
  int i;
  for (i = 21 - 1; i >= 0; i--) {
    if (lvl[x] - (1 << i) >= lvl[y]) {
      x = LCA[x][i];
    }
    if (lvl[y] - (1 << i) >= lvl[x]) {
      y = LCA[y][i];
    }
  }
  for (i = 21 - 1; i >= 0; i--) {
    if (LCA[x][i] != LCA[y][i]) {
      x = LCA[x][i];
      y = LCA[y][i];
    }
  }
  if (x != y) {
    x = LCA[x][0];
  }
  return x;
}
void calculate_prefixes(int vertex, int prv, int prv_edge) {
  int i;
  int sz = v[vertex].size();
  if (prv == -1) {
    prefix[vertex] = 0;
  } else {
    prefix[vertex] = prefix[prv] - prv_edge;
  }
  prefix[vertex] += dp[vertex];
  if (prv != -1) {
    long long aux = dp[vertex] - 2 * prv_edge;
    if (aux < 0) {
      aux = 0;
    }
    prefix[vertex] -= aux;
  }
  for (i = 0; i < sz; i++) {
    if (v[vertex][i].first == prv) {
      continue;
    }
    calculate_prefixes(v[vertex][i].first, vertex, v[vertex][i].second);
  }
}
void input() {
  scanf("%d%d", &n, &q);
  int i;
  for (i = 1; i <= n; i++) {
    scanf("%d", &value[i]);
  }
  for (i = 1; i < n; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    v[x].push_back(make_pair(y, z));
    v[y].push_back(make_pair(x, z));
  }
}
void solve() {
  dfs_bot(1, -1);
  dfs_top(1, -1, 0);
  calculate_prefixes(1, -1, 0);
  while (q != 0) {
    q--;
    int x, y;
    scanf("%d%d", &x, &y);
    int u = get_lca(x, y);
    long long ans = up[u];
    ans += prefix[x] - prefix[u];
    ans += prefix[y] - prefix[u];
    printf("%I64d\n", ans);
  }
}
int main() {
  input();
  solve();
  return 0;
}

```