#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 1;
const int maxlog = 20;
long long a[maxn];
vector<pair<int, pair<long long, long long> > > g[maxn];
long long d[maxn];
int h[maxn], p[maxn][maxlog];
void dfs(int cur) {
  for (int i = 1; i < maxlog; ++i) p[cur][i] = p[p[cur][i - 1]][i - 1];
  d[cur] = a[cur];
  for (auto &pr : g[cur]) {
    int nxt = pr.first;
    long long w = pr.second.first;
    if (nxt == p[cur][0]) continue;
    h[nxt] = h[cur] + 1;
    p[nxt][0] = cur;
    dfs(nxt);
    d[cur] += (pr.second.second = max(0ll, d[nxt] - 2 * w));
  }
}
long long e[maxn];
void dfs(int cur, long long tmp) {
  d[cur] += tmp;
  for (auto &pr : g[cur]) {
    int nxt = pr.first;
    long long w = pr.second.first;
    if (nxt == p[cur][0]) continue;
    long long tmp = max(0ll, d[cur] - pr.second.second - 2 * w);
    e[nxt] = e[cur] - w - pr.second.second - tmp;
    dfs(nxt, tmp);
  }
}
void dfs2(int cur) {
  for (auto &pr : g[cur]) {
    int nxt = pr.first;
    long long w = pr.second.first;
    if (nxt == p[cur][0]) continue;
    d[nxt] += d[cur];
    dfs2(nxt);
  }
}
int lca(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  int dif = h[u] - h[v];
  for (int i = 0; i < maxlog; ++i)
    if ((dif >> i) & 1) u = p[u][i];
  for (int i = maxlog - 1; i >= 0; --i)
    if (p[u][i] != p[v][i]) {
      u = p[u][i];
      v = p[v][i];
    }
  if (u == v)
    return u;
  else
    return p[u][0];
}
void print(int n) {
  for (int i = 1; i <= n; ++i) cerr << i << " = " << d[i] << endl;
  cerr << endl;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%lld", a + i);
  for (int i = 1; i <= n - 1; ++i) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    g[u].push_back(make_pair(v, make_pair(w, 0)));
    g[v].push_back(make_pair(u, make_pair(w, 0)));
  }
  dfs(1);
  dfs(1, 0);
  dfs2(1);
  for (int i = 0; i < q; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    int l = lca(u, v);
    int pl = p[l][0];
    long long sumd = d[u] + d[v] - d[l] - d[pl];
    long long sume = e[u] + e[v] - 2 * e[l];
    printf("%lld\n", sumd + sume);
  }
}

### Prompt

Generate a CPP solution to 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 maxn = 3e5 + 1;
const int maxlog = 20;
long long a[maxn];
vector<pair<int, pair<long long, long long> > > g[maxn];
long long d[maxn];
int h[maxn], p[maxn][maxlog];
void dfs(int cur) {
  for (int i = 1; i < maxlog; ++i) p[cur][i] = p[p[cur][i - 1]][i - 1];
  d[cur] = a[cur];
  for (auto &pr : g[cur]) {
    int nxt = pr.first;
    long long w = pr.second.first;
    if (nxt == p[cur][0]) continue;
    h[nxt] = h[cur] + 1;
    p[nxt][0] = cur;
    dfs(nxt);
    d[cur] += (pr.second.second = max(0ll, d[nxt] - 2 * w));
  }
}
long long e[maxn];
void dfs(int cur, long long tmp) {
  d[cur] += tmp;
  for (auto &pr : g[cur]) {
    int nxt = pr.first;
    long long w = pr.second.first;
    if (nxt == p[cur][0]) continue;
    long long tmp = max(0ll, d[cur] - pr.second.second - 2 * w);
    e[nxt] = e[cur] - w - pr.second.second - tmp;
    dfs(nxt, tmp);
  }
}
void dfs2(int cur) {
  for (auto &pr : g[cur]) {
    int nxt = pr.first;
    long long w = pr.second.first;
    if (nxt == p[cur][0]) continue;
    d[nxt] += d[cur];
    dfs2(nxt);
  }
}
int lca(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  int dif = h[u] - h[v];
  for (int i = 0; i < maxlog; ++i)
    if ((dif >> i) & 1) u = p[u][i];
  for (int i = maxlog - 1; i >= 0; --i)
    if (p[u][i] != p[v][i]) {
      u = p[u][i];
      v = p[v][i];
    }
  if (u == v)
    return u;
  else
    return p[u][0];
}
void print(int n) {
  for (int i = 1; i <= n; ++i) cerr << i << " = " << d[i] << endl;
  cerr << endl;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%lld", a + i);
  for (int i = 1; i <= n - 1; ++i) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    g[u].push_back(make_pair(v, make_pair(w, 0)));
    g[v].push_back(make_pair(u, make_pair(w, 0)));
  }
  dfs(1);
  dfs(1, 0);
  dfs2(1);
  for (int i = 0; i < q; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    int l = lca(u, v);
    int pl = p[l][0];
    long long sumd = d[u] + d[v] - d[l] - d[pl];
    long long sume = e[u] + e[v] - 2 * e[l];
    printf("%lld\n", sumd + sume);
  }
}

```