#include <bits/stdc++.h>
using namespace std;
const int _ = 3e5 + 55;
int N, Q, a[_];
int hd[_], nxt[_ << 1], to[_ << 1], val[_ << 1], tot(1);
long long vn[_], ve[_];
int fa[_][20], dep[_];
long long f[_], sf[_], g[_];
inline void addedge(int x, int y, int z) {
  nxt[++tot] = hd[x], hd[x] = tot, to[tot] = y, val[tot] = z;
}
void DFS(int u) {
  vn[u] = vn[fa[u][0]] + a[u], dep[u] = dep[fa[u][0]] + 1;
  for (int i(1), I(18); i <= I; ++i) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  for (int i(hd[u]), v(to[hd[u]]); i; v = to[i = nxt[i]])
    if (v != fa[u][0]) {
      ve[v] = ve[u] + val[i], fa[v][0] = u, DFS(v);
      f[v] = max(0ll, sf[v] + a[v] - 2 * val[i]);
      sf[u] += f[v];
    }
}
void DFS1(int u) {
  for (int i(hd[u]), v(to[hd[u]]); i; v = to[i = nxt[i]])
    if (v != fa[u][0]) {
      g[v] = max(0ll, g[u] + sf[u] - f[v] + a[u] - (2 * val[i]));
      DFS1(v);
    }
}
void DFS2(int u) {
  for (int i(hd[u]), v(to[hd[u]]); i; v = to[i = nxt[i]])
    if (v != fa[u][0]) {
      f[v] += f[u], sf[v] += sf[u], DFS2(v);
    }
}
int LCA(int x, int y) {
  if (dep[x] < dep[y]) x ^= y ^= x ^= y;
  for (int i(18), I(0); i >= I; --i)
    if (dep[fa[x][i]] > dep[y]) x = fa[x][i];
  if (dep[x] > dep[y]) x = fa[x][0];
  for (int i(18), I(0); i >= I; --i)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return x != y ? fa[x][0] : x;
}
int main() {
  scanf("%d%d", &N, &Q);
  for (int i(1), I(N); i <= I; ++i) scanf("%d", a + i);
  int x, y, z;
  for (int i(1), I(N - 1); i <= I; ++i)
    scanf("%d%d%d", &x, &y, &z), addedge(x, y, z), addedge(y, x, z);
  DFS(1);
  DFS1(1);
  DFS2(1);
  while (Q--) {
    scanf("%d%d", &x, &y);
    z = LCA(x, y);
    long long ans(vn[x] + vn[y] - vn[z] - vn[fa[z][0]] -
                  (ve[x] + ve[y] - 2 * ve[z]));
    ans += sf[x] + sf[y] - sf[z] - sf[fa[z][0]] + g[z];
    ans -= f[x] + f[y] - 2 * f[z];
    printf("%lld\n", ans);
  }
  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 _ = 3e5 + 55;
int N, Q, a[_];
int hd[_], nxt[_ << 1], to[_ << 1], val[_ << 1], tot(1);
long long vn[_], ve[_];
int fa[_][20], dep[_];
long long f[_], sf[_], g[_];
inline void addedge(int x, int y, int z) {
  nxt[++tot] = hd[x], hd[x] = tot, to[tot] = y, val[tot] = z;
}
void DFS(int u) {
  vn[u] = vn[fa[u][0]] + a[u], dep[u] = dep[fa[u][0]] + 1;
  for (int i(1), I(18); i <= I; ++i) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  for (int i(hd[u]), v(to[hd[u]]); i; v = to[i = nxt[i]])
    if (v != fa[u][0]) {
      ve[v] = ve[u] + val[i], fa[v][0] = u, DFS(v);
      f[v] = max(0ll, sf[v] + a[v] - 2 * val[i]);
      sf[u] += f[v];
    }
}
void DFS1(int u) {
  for (int i(hd[u]), v(to[hd[u]]); i; v = to[i = nxt[i]])
    if (v != fa[u][0]) {
      g[v] = max(0ll, g[u] + sf[u] - f[v] + a[u] - (2 * val[i]));
      DFS1(v);
    }
}
void DFS2(int u) {
  for (int i(hd[u]), v(to[hd[u]]); i; v = to[i = nxt[i]])
    if (v != fa[u][0]) {
      f[v] += f[u], sf[v] += sf[u], DFS2(v);
    }
}
int LCA(int x, int y) {
  if (dep[x] < dep[y]) x ^= y ^= x ^= y;
  for (int i(18), I(0); i >= I; --i)
    if (dep[fa[x][i]] > dep[y]) x = fa[x][i];
  if (dep[x] > dep[y]) x = fa[x][0];
  for (int i(18), I(0); i >= I; --i)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return x != y ? fa[x][0] : x;
}
int main() {
  scanf("%d%d", &N, &Q);
  for (int i(1), I(N); i <= I; ++i) scanf("%d", a + i);
  int x, y, z;
  for (int i(1), I(N - 1); i <= I; ++i)
    scanf("%d%d%d", &x, &y, &z), addedge(x, y, z), addedge(y, x, z);
  DFS(1);
  DFS1(1);
  DFS2(1);
  while (Q--) {
    scanf("%d%d", &x, &y);
    z = LCA(x, y);
    long long ans(vn[x] + vn[y] - vn[z] - vn[fa[z][0]] -
                  (ve[x] + ve[y] - 2 * ve[z]));
    ans += sf[x] + sf[y] - sf[z] - sf[fa[z][0]] + g[z];
    ans -= f[x] + f[y] - 2 * f[z];
    printf("%lld\n", ans);
  }
  return 0;
}

```