#include <bits/stdc++.h>
using namespace std;
long long a[300001], s[300001], t[300001];
struct edge {
  unsigned p, v;
  long long w, s;
};
struct query {
  unsigned u, v, w;
};
vector<edge> g[300001];
vector<query> qv;
vector<unsigned> qs[300001];
void adde(unsigned u, unsigned v, long long w) {
  g[u].push_back({(unsigned)g[v].size(), v, w, 0});
  g[v].push_back({(unsigned)g[u].size() - 1, u, w, 0});
}
void dfs1(unsigned u, unsigned f) {
  for (edge& e : g[u]) {
    if (e.v == f) continue;
    dfs1(e.v, u);
    s[u] += e.s = max(0ll, s[e.v] + a[e.v] - 2 * e.w);
  }
}
void dfs2(unsigned u, unsigned f) {
  for (edge e : g[u]) {
    if (e.v == f) continue;
    s[e.v] += g[e.v][e.p].s = max(0ll, s[u] + a[u] - 2 * e.w - e.s);
    t[e.v] = t[u] + s[e.v] - e.s - g[e.v][e.p].s + a[e.v] - e.w;
    dfs2(e.v, u);
  }
}
unsigned fs[300001];
bool vis[300001];
unsigned uf(unsigned x) { return fs[x] ? fs[x] = uf(fs[x]) : x; }
void dfs_lca(unsigned u, unsigned f) {
  vis[u] = 1;
  for (unsigned qi : qs[u]) {
    query& q = qv[qi];
    if (q.u != u) swap(q.u, q.v);
    if (vis[q.v]) q.w = uf(q.v);
  }
  for (edge e : g[u]) {
    if (e.v == f) continue;
    dfs_lca(e.v, u);
    fs[e.v] = u;
  }
}
int main(void) {
  unsigned n, q;
  scanf("%u%u", &n, &q);
  for (unsigned i = 1; i <= n; ++i) scanf("%u", a + i);
  for (unsigned i = 1; i < n; ++i) {
    unsigned u, v;
    long long w;
    scanf("%u%u%lld", &u, &v, &w);
    adde(u, v, w);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  for (unsigned i = 0; i != q; ++i) {
    unsigned u, v;
    scanf("%u%u", &u, &v);
    qs[u].push_back(qv.size());
    qs[v].push_back(qv.size());
    qv.push_back({u, v, 0});
  }
  dfs_lca(1, 0);
  for (unsigned i = 0; i != q; ++i) {
    unsigned u = qv[i].u, v = qv[i].v, w = qv[i].w;
    printf("%lld\n", t[u] + t[v] - 2 * t[w] + s[w] + a[w]);
  }
  return 0;
}

### 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;
long long a[300001], s[300001], t[300001];
struct edge {
  unsigned p, v;
  long long w, s;
};
struct query {
  unsigned u, v, w;
};
vector<edge> g[300001];
vector<query> qv;
vector<unsigned> qs[300001];
void adde(unsigned u, unsigned v, long long w) {
  g[u].push_back({(unsigned)g[v].size(), v, w, 0});
  g[v].push_back({(unsigned)g[u].size() - 1, u, w, 0});
}
void dfs1(unsigned u, unsigned f) {
  for (edge& e : g[u]) {
    if (e.v == f) continue;
    dfs1(e.v, u);
    s[u] += e.s = max(0ll, s[e.v] + a[e.v] - 2 * e.w);
  }
}
void dfs2(unsigned u, unsigned f) {
  for (edge e : g[u]) {
    if (e.v == f) continue;
    s[e.v] += g[e.v][e.p].s = max(0ll, s[u] + a[u] - 2 * e.w - e.s);
    t[e.v] = t[u] + s[e.v] - e.s - g[e.v][e.p].s + a[e.v] - e.w;
    dfs2(e.v, u);
  }
}
unsigned fs[300001];
bool vis[300001];
unsigned uf(unsigned x) { return fs[x] ? fs[x] = uf(fs[x]) : x; }
void dfs_lca(unsigned u, unsigned f) {
  vis[u] = 1;
  for (unsigned qi : qs[u]) {
    query& q = qv[qi];
    if (q.u != u) swap(q.u, q.v);
    if (vis[q.v]) q.w = uf(q.v);
  }
  for (edge e : g[u]) {
    if (e.v == f) continue;
    dfs_lca(e.v, u);
    fs[e.v] = u;
  }
}
int main(void) {
  unsigned n, q;
  scanf("%u%u", &n, &q);
  for (unsigned i = 1; i <= n; ++i) scanf("%u", a + i);
  for (unsigned i = 1; i < n; ++i) {
    unsigned u, v;
    long long w;
    scanf("%u%u%lld", &u, &v, &w);
    adde(u, v, w);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  for (unsigned i = 0; i != q; ++i) {
    unsigned u, v;
    scanf("%u%u", &u, &v);
    qs[u].push_back(qv.size());
    qs[v].push_back(qv.size());
    qv.push_back({u, v, 0});
  }
  dfs_lca(1, 0);
  for (unsigned i = 0; i != q; ++i) {
    unsigned u = qv[i].u, v = qv[i].v, w = qv[i].w;
    printf("%lld\n", t[u] + t[v] - 2 * t[w] + s[w] + a[w]);
  }
  return 0;
}

```