#include <bits/stdc++.h>
using namespace std;
const int MN = 3e5 + 5;
int to[MN << 1], nxt[MN << 1], c[MN << 1], h[MN], cnt = 1;
inline void ins(int s, int t, int w) {
  to[++cnt] = t;
  nxt[cnt] = h[s];
  h[s] = cnt;
  c[cnt] = w;
}
int n, q, a[MN];
int fa[MN], dep[MN], siz[MN], son[MN];
long long dis[MN], f[MN], F[MN], G[MN];
void dfs1(int st) {
  dep[st] = dep[fa[st]] + 1;
  siz[st] = 1;
  f[st] = a[st];
  for (register int i = h[st]; i; i = nxt[i])
    if (to[i] != fa[st]) {
      fa[to[i]] = st;
      dis[to[i]] = dis[st] + c[i];
      dfs1(to[i]);
      siz[st] += siz[to[i]];
      if (siz[son[st]] < siz[to[i]]) son[st] = to[i];
      f[st] += max(0ll, f[to[i]] - 2ll * c[i]);
    }
}
int top[MN];
void dfs2(int st, int tp) {
  top[st] = tp;
  if (son[st]) dfs2(son[st], tp);
  for (register int i = h[st]; i; i = nxt[i])
    if (to[i] != fa[st] && to[i] != son[st]) dfs2(to[i], to[i]);
}
void dfs3(int st) {
  for (register int i = h[st]; i; i = nxt[i])
    if (to[i] != fa[st]) {
      F[to[i]] = f[to[i]] + max(0ll, F[st] - max(2ll * c[i], f[to[i]]));
      G[to[i]] = G[st] + min(f[to[i]], 2ll * c[i]);
      dfs3(to[i]);
    }
}
inline int LCA(int x, int y) {
  while (top[x] ^ top[y])
    dep[top[x]] > dep[top[y]] ? x = fa[top[x]] : y = fa[top[y]];
  return dep[x] > dep[y] ? y : x;
}
int main() {
  scanf("%d%d", &n, &q);
  for (register int i = 1; i <= n; i++) scanf("%d", a + i);
  for (register int i = 1, s, t, c; i < n; i++) {
    scanf("%d%d%d", &s, &t, &c);
    ins(s, t, c);
    ins(t, s, c);
  }
  dfs1(1);
  dfs2(1, 1);
  F[1] = f[1];
  dfs3(1);
  while (q--) {
    static int s, t, lca;
    scanf("%d%d", &s, &t);
    lca = LCA(s, t);
    printf("%lld\n",
           2 * dis[lca] - dis[s] - dis[t] + F[lca] + G[s] + G[t] - 2 * G[lca]);
  }
  return 0;
}

### Prompt

Your task is to create 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 MN = 3e5 + 5;
int to[MN << 1], nxt[MN << 1], c[MN << 1], h[MN], cnt = 1;
inline void ins(int s, int t, int w) {
  to[++cnt] = t;
  nxt[cnt] = h[s];
  h[s] = cnt;
  c[cnt] = w;
}
int n, q, a[MN];
int fa[MN], dep[MN], siz[MN], son[MN];
long long dis[MN], f[MN], F[MN], G[MN];
void dfs1(int st) {
  dep[st] = dep[fa[st]] + 1;
  siz[st] = 1;
  f[st] = a[st];
  for (register int i = h[st]; i; i = nxt[i])
    if (to[i] != fa[st]) {
      fa[to[i]] = st;
      dis[to[i]] = dis[st] + c[i];
      dfs1(to[i]);
      siz[st] += siz[to[i]];
      if (siz[son[st]] < siz[to[i]]) son[st] = to[i];
      f[st] += max(0ll, f[to[i]] - 2ll * c[i]);
    }
}
int top[MN];
void dfs2(int st, int tp) {
  top[st] = tp;
  if (son[st]) dfs2(son[st], tp);
  for (register int i = h[st]; i; i = nxt[i])
    if (to[i] != fa[st] && to[i] != son[st]) dfs2(to[i], to[i]);
}
void dfs3(int st) {
  for (register int i = h[st]; i; i = nxt[i])
    if (to[i] != fa[st]) {
      F[to[i]] = f[to[i]] + max(0ll, F[st] - max(2ll * c[i], f[to[i]]));
      G[to[i]] = G[st] + min(f[to[i]], 2ll * c[i]);
      dfs3(to[i]);
    }
}
inline int LCA(int x, int y) {
  while (top[x] ^ top[y])
    dep[top[x]] > dep[top[y]] ? x = fa[top[x]] : y = fa[top[y]];
  return dep[x] > dep[y] ? y : x;
}
int main() {
  scanf("%d%d", &n, &q);
  for (register int i = 1; i <= n; i++) scanf("%d", a + i);
  for (register int i = 1, s, t, c; i < n; i++) {
    scanf("%d%d%d", &s, &t, &c);
    ins(s, t, c);
    ins(t, s, c);
  }
  dfs1(1);
  dfs2(1, 1);
  F[1] = f[1];
  dfs3(1);
  while (q--) {
    static int s, t, lca;
    scanf("%d%d", &s, &t);
    lca = LCA(s, t);
    printf("%lld\n",
           2 * dis[lca] - dis[s] - dis[t] + F[lca] + G[s] + G[t] - 2 * G[lca]);
  }
  return 0;
}

```