#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
long long num[N], dp1[N], dp2[N], sum[N];
int ne[N << 1], val[N], a[N], fi[N], zz[N << 1], sl[N << 1], fa[N], sd[N],
    Size[N], son[N], n, m, x, y, z, tot, top[N];
inline void jb(int x, int y, int z) {
  ne[++tot] = fi[x];
  fi[x] = tot;
  zz[tot] = y;
  sl[tot] = z;
}
void dfs1(int x, int y, int z) {
  fa[x] = y;
  sd[x] = z;
  Size[x] = 1;
  for (int i = fi[x]; i; i = ne[i])
    if (zz[i] != y) {
      val[zz[i]] = sl[i];
      dfs1(zz[i], x, z + 1);
      Size[x] += Size[zz[i]];
      if (Size[zz[i]] > Size[son[x]]) son[x] = zz[i];
    }
}
void dfs2(int x, int y) {
  top[x] = y;
  if (son[x]) dfs2(son[x], y);
  for (int i = fi[x]; i; i = ne[i])
    if (zz[i] != fa[x] && zz[i] != son[x]) dfs2(zz[i], zz[i]);
}
void solve1(int x) {
  for (int i = fi[x]; i; i = ne[i])
    if (zz[i] != fa[x]) {
      solve1(zz[i]);
      sum[x] += dp1[zz[i]];
    }
  dp1[x] = max(0ll, sum[x] + a[x] - 2 * val[x]);
}
void solve2(int x) {
  dp2[x] = max(0ll, dp2[fa[x]] + sum[fa[x]] - dp1[x] - 2 * val[x] + a[fa[x]]);
  num[x] = a[x] - val[x] + sum[fa[x]] - dp1[x] + num[fa[x]];
  for (int i = fi[x]; i; i = ne[i])
    if (zz[i] != fa[x]) solve2(zz[i]);
}
inline long long ask(int x, int y) {
  long long ans = sum[x] + sum[y];
  int fx = top[x], fy = top[y];
  while (fx != fy) {
    if (sd[fx] < sd[fy]) {
      swap(x, y);
      swap(fx, fy);
    }
    ans += num[x] - num[fx];
    ans += a[fx] - val[fx] + sum[fa[fx]] - dp1[fx];
    x = fa[fx];
    fx = top[x];
  }
  if (sd[x] > sd[y]) swap(x, y);
  ans += num[y] - num[x];
  ans += dp2[x] - sum[x] + a[x];
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    scanf("%d%d%d", &x, &y, &z);
    jb(x, y, z);
    jb(y, x, z);
  }
  dfs1(1, 0, 0);
  dfs2(1, 1);
  solve1(1);
  solve2(1);
  while (m--) {
    scanf("%d%d", &x, &y);
    printf("%lld\n", ask(x, y));
  }
}

### Prompt

In CPP, your task is to solve 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 N = 300005;
long long num[N], dp1[N], dp2[N], sum[N];
int ne[N << 1], val[N], a[N], fi[N], zz[N << 1], sl[N << 1], fa[N], sd[N],
    Size[N], son[N], n, m, x, y, z, tot, top[N];
inline void jb(int x, int y, int z) {
  ne[++tot] = fi[x];
  fi[x] = tot;
  zz[tot] = y;
  sl[tot] = z;
}
void dfs1(int x, int y, int z) {
  fa[x] = y;
  sd[x] = z;
  Size[x] = 1;
  for (int i = fi[x]; i; i = ne[i])
    if (zz[i] != y) {
      val[zz[i]] = sl[i];
      dfs1(zz[i], x, z + 1);
      Size[x] += Size[zz[i]];
      if (Size[zz[i]] > Size[son[x]]) son[x] = zz[i];
    }
}
void dfs2(int x, int y) {
  top[x] = y;
  if (son[x]) dfs2(son[x], y);
  for (int i = fi[x]; i; i = ne[i])
    if (zz[i] != fa[x] && zz[i] != son[x]) dfs2(zz[i], zz[i]);
}
void solve1(int x) {
  for (int i = fi[x]; i; i = ne[i])
    if (zz[i] != fa[x]) {
      solve1(zz[i]);
      sum[x] += dp1[zz[i]];
    }
  dp1[x] = max(0ll, sum[x] + a[x] - 2 * val[x]);
}
void solve2(int x) {
  dp2[x] = max(0ll, dp2[fa[x]] + sum[fa[x]] - dp1[x] - 2 * val[x] + a[fa[x]]);
  num[x] = a[x] - val[x] + sum[fa[x]] - dp1[x] + num[fa[x]];
  for (int i = fi[x]; i; i = ne[i])
    if (zz[i] != fa[x]) solve2(zz[i]);
}
inline long long ask(int x, int y) {
  long long ans = sum[x] + sum[y];
  int fx = top[x], fy = top[y];
  while (fx != fy) {
    if (sd[fx] < sd[fy]) {
      swap(x, y);
      swap(fx, fy);
    }
    ans += num[x] - num[fx];
    ans += a[fx] - val[fx] + sum[fa[fx]] - dp1[fx];
    x = fa[fx];
    fx = top[x];
  }
  if (sd[x] > sd[y]) swap(x, y);
  ans += num[y] - num[x];
  ans += dp2[x] - sum[x] + a[x];
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    scanf("%d%d%d", &x, &y, &z);
    jb(x, y, z);
    jb(y, x, z);
  }
  dfs1(1, 0, 0);
  dfs2(1, 1);
  solve1(1);
  solve2(1);
  while (m--) {
    scanf("%d%d", &x, &y);
    printf("%lld\n", ask(x, y));
  }
}

```