#include <bits/stdc++.h>
using namespace std;
const int maxn = 300010;
int n, q, a[maxn], fa[maxn], dep[maxn], sz[maxn], son[maxn], top[maxn],
    tmp[maxn];
long long up[maxn], down[maxn], dist[maxn];
vector<int> G[maxn], W[maxn];
void dfs1(int v, int f) {
  fa[v] = f, dep[v] = dep[f] + 1, sz[v] = 1;
  for (int i = 0, u; i < G[v].size(); i++) {
    if ((u = G[v][i]) == f) continue;
    dfs1(u, v), sz[v] += sz[u];
    if (sz[u] > sz[son[v]]) son[v] = u;
    down[v] = max(down[v], down[v] + down[u] + a[u] - 2 * W[v][i]);
  }
}
void dfs2(int v, int rt, int w) {
  top[v] = rt, dist[v] = dist[fa[v]] + a[v] - (tmp[v] = w) + down[v];
  dist[v] -= max(0LL, down[v] + a[v] - 2 * w);
  for (int i = 0, u; i < G[v].size(); i++) {
    if ((u = G[v][i]) == fa[v]) continue;
    long long x = down[u] + a[u] - 2 * W[v][i], y = up[v] + a[v] - 2 * W[v][i];
    up[u] = max(0LL, down[v] - max(0LL, x) + y);
    u ^ son[v] ? dfs2(u, u, W[v][i]) : dfs2(u, rt, W[v][i]);
  }
}
int lca(int u, int v) {
  for (; top[u] ^ top[v]; u = fa[top[u]]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
  }
  return dep[u] < dep[v] ? u : v;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[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);
  }
  dfs1(1, 0), dfs2(1, 1, 0);
  for (int i = 1, u, v, p; i <= q; i++) {
    scanf("%d %d", &u, &v), p = lca(u, v);
    long long ans = dist[u] + dist[v] - dist[p] - dist[fa[p]] + up[p];
    ans += tmp[p] + max(0LL, down[p] + a[p] - 2LL * tmp[p]);
    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 maxn = 300010;
int n, q, a[maxn], fa[maxn], dep[maxn], sz[maxn], son[maxn], top[maxn],
    tmp[maxn];
long long up[maxn], down[maxn], dist[maxn];
vector<int> G[maxn], W[maxn];
void dfs1(int v, int f) {
  fa[v] = f, dep[v] = dep[f] + 1, sz[v] = 1;
  for (int i = 0, u; i < G[v].size(); i++) {
    if ((u = G[v][i]) == f) continue;
    dfs1(u, v), sz[v] += sz[u];
    if (sz[u] > sz[son[v]]) son[v] = u;
    down[v] = max(down[v], down[v] + down[u] + a[u] - 2 * W[v][i]);
  }
}
void dfs2(int v, int rt, int w) {
  top[v] = rt, dist[v] = dist[fa[v]] + a[v] - (tmp[v] = w) + down[v];
  dist[v] -= max(0LL, down[v] + a[v] - 2 * w);
  for (int i = 0, u; i < G[v].size(); i++) {
    if ((u = G[v][i]) == fa[v]) continue;
    long long x = down[u] + a[u] - 2 * W[v][i], y = up[v] + a[v] - 2 * W[v][i];
    up[u] = max(0LL, down[v] - max(0LL, x) + y);
    u ^ son[v] ? dfs2(u, u, W[v][i]) : dfs2(u, rt, W[v][i]);
  }
}
int lca(int u, int v) {
  for (; top[u] ^ top[v]; u = fa[top[u]]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
  }
  return dep[u] < dep[v] ? u : v;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[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);
  }
  dfs1(1, 0), dfs2(1, 1, 0);
  for (int i = 1, u, v, p; i <= q; i++) {
    scanf("%d %d", &u, &v), p = lca(u, v);
    long long ans = dist[u] + dist[v] - dist[p] - dist[fa[p]] + up[p];
    ans += tmp[p] + max(0LL, down[p] + a[p] - 2LL * tmp[p]);
    printf("%lld\n", ans);
  }
  return 0;
}

```