#include <bits/stdc++.h>
using namespace std;
int n, q;
long long a[300005];
vector<pair<int, int> > adj[300005];
long long subtree[300005];
long long tree[300005];
long long ctree[300005];
void go(int u, int p) {
  subtree[u] = a[u];
  for (auto e : adj[u]) {
    int v = e.first;
    long long w = e.second;
    if (v == p) continue;
    go(v, u);
    ctree[v] = max(0LL, subtree[v] - 2 * w);
    subtree[u] += ctree[v];
  }
}
void go(int u, int p, long long pc) {
  tree[u] = subtree[u] + pc;
  for (auto e : adj[u]) {
    int v = e.first;
    long long w = e.second;
    if (v == p) continue;
    long long nc = tree[u] - ctree[v];
    nc = max(0LL, nc - 2 * w);
    ctree[v] += w;
    go(v, u, nc);
  }
}
long long subtreesum[300005][25];
long long ctreesum[300005][25];
long long par[300005][25];
int level[300005];
void build(int u, int p) {
  level[u] = 1 + level[p];
  par[u][0] = p;
  for (int i = 1; i <= 20; i++) par[u][i] = par[par[u][i - 1]][i - 1];
  subtreesum[u][0] = subtree[u];
  for (int i = 1; i <= 20; i++)
    subtreesum[u][i] = subtreesum[u][i - 1] + subtreesum[par[u][i - 1]][i - 1];
  ctreesum[u][0] = ctree[u];
  for (int i = 1; i <= 20; i++)
    ctreesum[u][i] = ctreesum[u][i - 1] + ctreesum[par[u][i - 1]][i - 1];
  for (auto e : adj[u]) {
    int v = e.first;
    long long w = e.second;
    if (v == p) continue;
    build(v, u);
  }
}
int lca(int u, int v) {
  if (level[u] < level[v]) swap(u, v);
  int tmp = 1;
  for (; (1 << tmp) <= level[u]; ++tmp)
    ;
  --tmp;
  for (int i = tmp; i >= 0; --i) {
    if (level[u] - (1 << i) >= level[v]) u = par[u][i];
  }
  if (u == v) return u;
  for (int i = tmp; i >= 0; --i)
    if (par[u][i] != par[v][i]) u = par[u][i], v = par[v][i];
  return par[u][0];
}
long long pathQuery(int u, int p) {
  long long ret = 0;
  for (int i = 20; i >= 0; i--) {
    if (u == p) break;
    if (level[par[u][i]] < level[p]) continue;
    ret += subtreesum[u][i] - ctreesum[u][i];
    u = par[u][i];
  }
  return ret;
}
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    long long w;
    scanf("%d %d %lld", &u, &v, &w);
    adj[u].push_back({v, w});
    adj[v].push_back({u, w});
  }
  go(1, 0);
  go(1, 0, 0);
  build(1, 0);
  for (int i = 0; i < q; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    int lc = lca(u, v);
    long long ret = pathQuery(u, lc);
    ret += pathQuery(v, lc);
    ret += tree[lc];
    cout << ret << endl;
  }
  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;
int n, q;
long long a[300005];
vector<pair<int, int> > adj[300005];
long long subtree[300005];
long long tree[300005];
long long ctree[300005];
void go(int u, int p) {
  subtree[u] = a[u];
  for (auto e : adj[u]) {
    int v = e.first;
    long long w = e.second;
    if (v == p) continue;
    go(v, u);
    ctree[v] = max(0LL, subtree[v] - 2 * w);
    subtree[u] += ctree[v];
  }
}
void go(int u, int p, long long pc) {
  tree[u] = subtree[u] + pc;
  for (auto e : adj[u]) {
    int v = e.first;
    long long w = e.second;
    if (v == p) continue;
    long long nc = tree[u] - ctree[v];
    nc = max(0LL, nc - 2 * w);
    ctree[v] += w;
    go(v, u, nc);
  }
}
long long subtreesum[300005][25];
long long ctreesum[300005][25];
long long par[300005][25];
int level[300005];
void build(int u, int p) {
  level[u] = 1 + level[p];
  par[u][0] = p;
  for (int i = 1; i <= 20; i++) par[u][i] = par[par[u][i - 1]][i - 1];
  subtreesum[u][0] = subtree[u];
  for (int i = 1; i <= 20; i++)
    subtreesum[u][i] = subtreesum[u][i - 1] + subtreesum[par[u][i - 1]][i - 1];
  ctreesum[u][0] = ctree[u];
  for (int i = 1; i <= 20; i++)
    ctreesum[u][i] = ctreesum[u][i - 1] + ctreesum[par[u][i - 1]][i - 1];
  for (auto e : adj[u]) {
    int v = e.first;
    long long w = e.second;
    if (v == p) continue;
    build(v, u);
  }
}
int lca(int u, int v) {
  if (level[u] < level[v]) swap(u, v);
  int tmp = 1;
  for (; (1 << tmp) <= level[u]; ++tmp)
    ;
  --tmp;
  for (int i = tmp; i >= 0; --i) {
    if (level[u] - (1 << i) >= level[v]) u = par[u][i];
  }
  if (u == v) return u;
  for (int i = tmp; i >= 0; --i)
    if (par[u][i] != par[v][i]) u = par[u][i], v = par[v][i];
  return par[u][0];
}
long long pathQuery(int u, int p) {
  long long ret = 0;
  for (int i = 20; i >= 0; i--) {
    if (u == p) break;
    if (level[par[u][i]] < level[p]) continue;
    ret += subtreesum[u][i] - ctreesum[u][i];
    u = par[u][i];
  }
  return ret;
}
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    long long w;
    scanf("%d %d %lld", &u, &v, &w);
    adj[u].push_back({v, w});
    adj[v].push_back({u, w});
  }
  go(1, 0);
  go(1, 0, 0);
  build(1, 0);
  for (int i = 0; i < q; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    int lc = lca(u, v);
    long long ret = pathQuery(u, lc);
    ret += pathQuery(v, lc);
    ret += tree[lc];
    cout << ret << endl;
  }
  return 0;
}

```