#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int n, q, w[N], hs[N], fa[N], tp[N], sz[N], d[N];
long long dn[N], edg[N], s[N], up[N], v[N];
vector<int> G[N], W[N];
void dfs(int x) {
  dn[x] = w[x];
  d[x] = d[fa[x]] + 1;
  sz[x] = 1;
  for (int i = 0, y; i < G[x].size(); i++)
    if ((y = G[x][i]) != fa[x]) {
      fa[y] = x;
      v[y] = W[x][i];
      dfs(y);
      sz[x] += sz[y];
      dn[x] += edg[y];
      hs[x] = (sz[y] > sz[hs[x]]) ? y : hs[x];
    }
  edg[x] = max(0ll, dn[x] - 2 * v[x]);
}
void dfss(int x) {
  s[x] = dn[fa[x]] - edg[x] - v[x];
  s[x] += s[fa[x]];
  up[x] = max(0ll, up[fa[x]] + dn[fa[x]] - edg[x] - 2 * v[x]);
  if (hs[x]) tp[hs[x]] = tp[x], dfss(hs[x]);
  for (int i = 0, y; i < G[x].size(); i++)
    if (G[x][i] != fa[x] && G[x][i] != hs[x])
      tp[G[x][i]] = G[x][i], dfss(G[x][i]);
}
inline int lca(int x, int y) {
  for (; tp[x] != tp[y]; x = fa[tp[x]])
    if (d[tp[x]] < d[tp[y]]) swap(x, y);
  return d[x] < d[y] ? x : y;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  for (int i = 1, u, v, w; i < n; i++) {
    scanf("%d%d%d", &u, &v, &w);
    G[u].push_back(v);
    W[u].push_back(w);
    G[v].push_back(u);
    W[v].push_back(w);
  }
  dfs(1);
  tp[1] = 1;
  dfss(1);
  for (int i = 1, u, v, t; i <= q; i++) {
    scanf("%d%d", &u, &v);
    t = lca(u, v);
    printf("%lld\n", s[u] + s[v] - (s[t] << 1) + dn[u] + dn[v] - dn[t] + up[t]);
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 = 300005;
int n, q, w[N], hs[N], fa[N], tp[N], sz[N], d[N];
long long dn[N], edg[N], s[N], up[N], v[N];
vector<int> G[N], W[N];
void dfs(int x) {
  dn[x] = w[x];
  d[x] = d[fa[x]] + 1;
  sz[x] = 1;
  for (int i = 0, y; i < G[x].size(); i++)
    if ((y = G[x][i]) != fa[x]) {
      fa[y] = x;
      v[y] = W[x][i];
      dfs(y);
      sz[x] += sz[y];
      dn[x] += edg[y];
      hs[x] = (sz[y] > sz[hs[x]]) ? y : hs[x];
    }
  edg[x] = max(0ll, dn[x] - 2 * v[x]);
}
void dfss(int x) {
  s[x] = dn[fa[x]] - edg[x] - v[x];
  s[x] += s[fa[x]];
  up[x] = max(0ll, up[fa[x]] + dn[fa[x]] - edg[x] - 2 * v[x]);
  if (hs[x]) tp[hs[x]] = tp[x], dfss(hs[x]);
  for (int i = 0, y; i < G[x].size(); i++)
    if (G[x][i] != fa[x] && G[x][i] != hs[x])
      tp[G[x][i]] = G[x][i], dfss(G[x][i]);
}
inline int lca(int x, int y) {
  for (; tp[x] != tp[y]; x = fa[tp[x]])
    if (d[tp[x]] < d[tp[y]]) swap(x, y);
  return d[x] < d[y] ? x : y;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  for (int i = 1, u, v, w; i < n; i++) {
    scanf("%d%d%d", &u, &v, &w);
    G[u].push_back(v);
    W[u].push_back(w);
    G[v].push_back(u);
    W[v].push_back(w);
  }
  dfs(1);
  tp[1] = 1;
  dfss(1);
  for (int i = 1, u, v, t; i <= q; i++) {
    scanf("%d%d", &u, &v);
    t = lca(u, v);
    printf("%lld\n", s[u] + s[v] - (s[t] << 1) + dn[u] + dn[v] - dn[t] + up[t]);
  }
  return 0;
}

```