#include <bits/stdc++.h>
using namespace std;
struct edge {
  int from, to, dist;
};
vector<edge> edges;
const int maxn = 4e5 + 233;
int f[maxn][21];
long long dp[maxn][3];
long long v_sum[maxn], e_sum[maxn];
long long sum[maxn][3];
int d[maxn];
vector<int> G[maxn];
int n, q;
int a[maxn];
void addedge(int from, int to, int dist) {
  edges.push_back(edge{from, to, dist});
  G[from].emplace_back(edges.size() - 1);
}
int lca(int x, int y) {
  if (d[x] > d[y]) swap(x, y);
  for (int i = 20; i >= 0; i--) {
    if (d[f[y][i]] >= d[x]) y = f[y][i];
  }
  if (x == y) return x;
  for (int i = 20; i >= 0; i--) {
    if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
  }
  return f[x][0];
}
void dfs1(int u, int fa) {
  d[u] = d[fa] + 1;
  f[u][0] = fa;
  v_sum[u] = v_sum[fa] + a[u];
  dp[u][0] = a[u];
  for (int i = 1; i <= 20; i++) {
    f[u][i] = f[f[u][i - 1]][i - 1];
  }
  for (int i = 0; i < G[u].size(); i++) {
    edge &e = edges[G[u][i]];
    if (e.to == fa) continue;
    e_sum[e.to] = e_sum[u] + e.dist;
    dfs1(e.to, u);
    dp[u][0] += max(0ll, dp[e.to][0] - 2 * e.dist);
  }
}
void dfs2(int u, int fa, long long t1, long long t2) {
  long long r1 = 0, r2 = a[u];
  dp[u][1] = t1;
  dp[u][2] = t2;
  for (int i = 0; i < 3; i++) {
    sum[u][i] = sum[fa][i] + dp[u][i];
  }
  for (int i = 0; i < G[u].size(); i++) {
    edge &e = edges[G[u][i]];
    if (e.to == fa) continue;
    r1 += max(0ll, dp[e.to][0] - 2 * e.dist);
    r2 += max(0ll, dp[e.to][0] - 2 * e.dist);
  }
  for (int i = 0; i < G[u].size(); i++) {
    edge &e = edges[G[u][i]];
    if (e.to == fa) continue;
    r1 -= max(0ll, dp[e.to][0] - 2 * e.dist);
    r2 -= max(0ll, dp[e.to][0] - 2 * e.dist);
    dfs2(e.to, u, max(0ll, r1), max(0ll, r2 + t2 - 2 * e.dist));
    r1 += max(0ll, dp[e.to][0] - 2 * e.dist);
    r2 += max(0ll, dp[e.to][0] - 2 * e.dist);
  }
}
long long cal(int x, int y) {
  return sum[x][1] - sum[y][1] + dp[x][0] + dp[y][2] + v_sum[f[x][0]] -
         v_sum[f[y][0]] - (e_sum[x] - e_sum[y]);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    addedge(u, v, w);
    addedge(v, u, w);
  }
  dfs1(1, 0);
  dfs2(1, 0, 0, 0);
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (d[x] < d[y]) swap(x, y);
    int lc = lca(x, y);
    long long res = 0;
    if (y != lc) {
      res = cal(x, lc) + cal(y, lc);
      res -= dp[lc][2] + a[lc];
      for (int i = 0; i < G[lc].size(); i++) {
        edge &e = edges[G[lc][i]];
        if (e.to == f[lc][0]) continue;
        res -= max(0ll, dp[e.to][0] - 2 * e.dist);
      }
    } else
      res = cal(x, lc);
    cout << res << "\n";
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
struct edge {
  int from, to, dist;
};
vector<edge> edges;
const int maxn = 4e5 + 233;
int f[maxn][21];
long long dp[maxn][3];
long long v_sum[maxn], e_sum[maxn];
long long sum[maxn][3];
int d[maxn];
vector<int> G[maxn];
int n, q;
int a[maxn];
void addedge(int from, int to, int dist) {
  edges.push_back(edge{from, to, dist});
  G[from].emplace_back(edges.size() - 1);
}
int lca(int x, int y) {
  if (d[x] > d[y]) swap(x, y);
  for (int i = 20; i >= 0; i--) {
    if (d[f[y][i]] >= d[x]) y = f[y][i];
  }
  if (x == y) return x;
  for (int i = 20; i >= 0; i--) {
    if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
  }
  return f[x][0];
}
void dfs1(int u, int fa) {
  d[u] = d[fa] + 1;
  f[u][0] = fa;
  v_sum[u] = v_sum[fa] + a[u];
  dp[u][0] = a[u];
  for (int i = 1; i <= 20; i++) {
    f[u][i] = f[f[u][i - 1]][i - 1];
  }
  for (int i = 0; i < G[u].size(); i++) {
    edge &e = edges[G[u][i]];
    if (e.to == fa) continue;
    e_sum[e.to] = e_sum[u] + e.dist;
    dfs1(e.to, u);
    dp[u][0] += max(0ll, dp[e.to][0] - 2 * e.dist);
  }
}
void dfs2(int u, int fa, long long t1, long long t2) {
  long long r1 = 0, r2 = a[u];
  dp[u][1] = t1;
  dp[u][2] = t2;
  for (int i = 0; i < 3; i++) {
    sum[u][i] = sum[fa][i] + dp[u][i];
  }
  for (int i = 0; i < G[u].size(); i++) {
    edge &e = edges[G[u][i]];
    if (e.to == fa) continue;
    r1 += max(0ll, dp[e.to][0] - 2 * e.dist);
    r2 += max(0ll, dp[e.to][0] - 2 * e.dist);
  }
  for (int i = 0; i < G[u].size(); i++) {
    edge &e = edges[G[u][i]];
    if (e.to == fa) continue;
    r1 -= max(0ll, dp[e.to][0] - 2 * e.dist);
    r2 -= max(0ll, dp[e.to][0] - 2 * e.dist);
    dfs2(e.to, u, max(0ll, r1), max(0ll, r2 + t2 - 2 * e.dist));
    r1 += max(0ll, dp[e.to][0] - 2 * e.dist);
    r2 += max(0ll, dp[e.to][0] - 2 * e.dist);
  }
}
long long cal(int x, int y) {
  return sum[x][1] - sum[y][1] + dp[x][0] + dp[y][2] + v_sum[f[x][0]] -
         v_sum[f[y][0]] - (e_sum[x] - e_sum[y]);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    addedge(u, v, w);
    addedge(v, u, w);
  }
  dfs1(1, 0);
  dfs2(1, 0, 0, 0);
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (d[x] < d[y]) swap(x, y);
    int lc = lca(x, y);
    long long res = 0;
    if (y != lc) {
      res = cal(x, lc) + cal(y, lc);
      res -= dp[lc][2] + a[lc];
      for (int i = 0; i < G[lc].size(); i++) {
        edge &e = edges[G[lc][i]];
        if (e.to == f[lc][0]) continue;
        res -= max(0ll, dp[e.to][0] - 2 * e.dist);
      }
    } else
      res = cal(x, lc);
    cout << res << "\n";
  }
  return 0;
}

```