#include <bits/stdc++.h>
inline int read() {
  int x = 0;
  char ch = getchar();
  for (; ch < 48 || ch > 57; ch = getchar())
    ;
  for (; ch > 47 && ch < 58; ch = getchar())
    x = (x << 1) + (x << 3) + (ch ^ 48);
  return x;
}
using namespace std;
int poi[300005], fir[300005], son[300005], fat[300005], top[300005],
    siz[300005], dep[300005], arr[300005 << 1], nex[300005 << 1],
    cot[300005 << 1], e;
long long dwn[300005], edg[300005], sum[300005], fly[300005], dis[300005];
void dfs(int u) {
  dwn[u] = poi[u];
  dep[u] = dep[fat[u]] + (siz[u] = 1);
  for (int i = fir[u], v; i; i = nex[i])
    if ((v = arr[i]) != fat[u]) {
      fat[v] = u;
      dis[v] = cot[i];
      dfs(v);
      siz[u] += siz[v];
      dwn[u] += edg[v];
      if (siz[v] > siz[son[u]]) son[u] = v;
    }
  edg[u] = max(0ll, dwn[u] - 2 * dis[u]);
}
void DFS(int u) {
  fly[u] =
      max(0ll, fly[fat[u]] + (sum[u] = dwn[fat[u]] - edg[u] - dis[u]) - dis[u]);
  sum[u] += sum[fat[u]];
  if (son[u]) {
    top[son[u]] = top[u], DFS(son[u]);
    for (int i = fir[u]; i; i = nex[i])
      if (arr[i] != fat[u] && arr[i] != son[u])
        top[arr[i]] = arr[i], DFS(arr[i]);
  }
}
int lca(int x, int y) {
  for (; top[x] != top[y];)
    dep[top[x]] > dep[top[y]] ? x = fat[top[x]] : y = fat[top[y]];
  return dep[x] < dep[y] ? x : y;
}
int main() {
  int n = read(), q = read();
  for (int i = 1; i <= n; ++i) poi[i] = read();
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read(), w = read();
    arr[++e] = v;
    nex[e] = fir[u];
    fir[u] = e;
    cot[e] = w;
    arr[++e] = u;
    nex[e] = fir[v];
    fir[v] = e;
    cot[e] = w;
  }
  dfs(1);
  top[1] = 1;
  DFS(1);
  for (int i = 0; i < q; ++i) {
    int u = read(), v = read(), l = lca(u, v);
    printf("%lld\n",
           sum[u] + sum[v] - (sum[l] << 1) + dwn[u] + dwn[v] - dwn[l] + fly[l]);
  }
  return 0;
}

### 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>
inline int read() {
  int x = 0;
  char ch = getchar();
  for (; ch < 48 || ch > 57; ch = getchar())
    ;
  for (; ch > 47 && ch < 58; ch = getchar())
    x = (x << 1) + (x << 3) + (ch ^ 48);
  return x;
}
using namespace std;
int poi[300005], fir[300005], son[300005], fat[300005], top[300005],
    siz[300005], dep[300005], arr[300005 << 1], nex[300005 << 1],
    cot[300005 << 1], e;
long long dwn[300005], edg[300005], sum[300005], fly[300005], dis[300005];
void dfs(int u) {
  dwn[u] = poi[u];
  dep[u] = dep[fat[u]] + (siz[u] = 1);
  for (int i = fir[u], v; i; i = nex[i])
    if ((v = arr[i]) != fat[u]) {
      fat[v] = u;
      dis[v] = cot[i];
      dfs(v);
      siz[u] += siz[v];
      dwn[u] += edg[v];
      if (siz[v] > siz[son[u]]) son[u] = v;
    }
  edg[u] = max(0ll, dwn[u] - 2 * dis[u]);
}
void DFS(int u) {
  fly[u] =
      max(0ll, fly[fat[u]] + (sum[u] = dwn[fat[u]] - edg[u] - dis[u]) - dis[u]);
  sum[u] += sum[fat[u]];
  if (son[u]) {
    top[son[u]] = top[u], DFS(son[u]);
    for (int i = fir[u]; i; i = nex[i])
      if (arr[i] != fat[u] && arr[i] != son[u])
        top[arr[i]] = arr[i], DFS(arr[i]);
  }
}
int lca(int x, int y) {
  for (; top[x] != top[y];)
    dep[top[x]] > dep[top[y]] ? x = fat[top[x]] : y = fat[top[y]];
  return dep[x] < dep[y] ? x : y;
}
int main() {
  int n = read(), q = read();
  for (int i = 1; i <= n; ++i) poi[i] = read();
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read(), w = read();
    arr[++e] = v;
    nex[e] = fir[u];
    fir[u] = e;
    cot[e] = w;
    arr[++e] = u;
    nex[e] = fir[v];
    fir[v] = e;
    cot[e] = w;
  }
  dfs(1);
  top[1] = 1;
  DFS(1);
  for (int i = 0; i < q; ++i) {
    int u = read(), v = read(), l = lca(u, v);
    printf("%lld\n",
           sum[u] + sum[v] - (sum[l] << 1) + dwn[u] + dwn[v] - dwn[l] + fly[l]);
  }
  return 0;
}

```