#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 1;
int n, q;
int par[MAXN][20];
int a[MAXN];
int dep[MAXN];
long long len[MAXN];
long long sump[MAXN];
long long sums[MAXN];
struct _edge {
  int x, w;
  _edge(int x, int w) : x(x), w(w) {}
};
vector<_edge> edge[MAXN];
void dfs(int x, int p) {
  dep[x] = dep[p] + 1;
  sump[x] = a[x];
  for (_edge i : edge[x]) {
    if (i.x == p) continue;
    len[i.x] = len[x] + i.w;
    dfs(i.x, x);
    long long dis = max(0ll, sump[i.x] - (i.w << 1));
    sump[x] += dis;
    sums[i.x] += dis;
  }
}
void dfs2(int x, int p, long long v) {
  sump[x] += v;
  sums[x] += v;
  for (_edge i : edge[x]) {
    if (i.x == p) continue;
    long long dis = max(0ll, sump[i.x] - (i.w << 1));
    dfs2(i.x, x, max(0ll, sump[x] - dis - (i.w << 1)));
  }
}
void dfs3(int x, int p) {
  sump[x] += sump[p];
  sums[x] += sums[p];
  par[x][0] = p;
  for (int i = 1; i < 20; ++i) {
    par[x][i] = par[par[x][i - 1]][i - 1];
  }
  for (_edge i : edge[x]) {
    if (i.x == p) continue;
    dfs3(i.x, x);
  }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = 20; i--;) {
    if ((dep[x] - dep[y]) >> i) x = par[x][i];
  }
  if (x == y) return x;
  for (int i = 20; i--;) {
    if (par[x][i] != par[y][i]) {
      x = par[x][i];
      y = par[y][i];
    }
  }
  return par[x][0];
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 1; i < n; ++i) {
    int x, y, w;
    scanf("%d%d%d", &x, &y, &w);
    edge[x].emplace_back(y, w);
    edge[y].emplace_back(x, w);
  }
  dfs(1, 0);
  dfs2(1, 0, 0);
  dfs3(1, 0);
  while (q--) {
    int x, y, p;
    scanf("%d%d", &x, &y);
    p = lca(x, y);
    long long vs = sump[x] + sump[y] - sump[par[p][0]] - sump[p];
    long long es = sums[x] + sums[y] - (sums[p] << 1);
    long long ls = len[x] + len[y] - (len[p] << 1);
    printf("%lld\n", vs - es - ls);
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
int n, q;
int par[MAXN][20];
int a[MAXN];
int dep[MAXN];
long long len[MAXN];
long long sump[MAXN];
long long sums[MAXN];
struct _edge {
  int x, w;
  _edge(int x, int w) : x(x), w(w) {}
};
vector<_edge> edge[MAXN];
void dfs(int x, int p) {
  dep[x] = dep[p] + 1;
  sump[x] = a[x];
  for (_edge i : edge[x]) {
    if (i.x == p) continue;
    len[i.x] = len[x] + i.w;
    dfs(i.x, x);
    long long dis = max(0ll, sump[i.x] - (i.w << 1));
    sump[x] += dis;
    sums[i.x] += dis;
  }
}
void dfs2(int x, int p, long long v) {
  sump[x] += v;
  sums[x] += v;
  for (_edge i : edge[x]) {
    if (i.x == p) continue;
    long long dis = max(0ll, sump[i.x] - (i.w << 1));
    dfs2(i.x, x, max(0ll, sump[x] - dis - (i.w << 1)));
  }
}
void dfs3(int x, int p) {
  sump[x] += sump[p];
  sums[x] += sums[p];
  par[x][0] = p;
  for (int i = 1; i < 20; ++i) {
    par[x][i] = par[par[x][i - 1]][i - 1];
  }
  for (_edge i : edge[x]) {
    if (i.x == p) continue;
    dfs3(i.x, x);
  }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = 20; i--;) {
    if ((dep[x] - dep[y]) >> i) x = par[x][i];
  }
  if (x == y) return x;
  for (int i = 20; i--;) {
    if (par[x][i] != par[y][i]) {
      x = par[x][i];
      y = par[y][i];
    }
  }
  return par[x][0];
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 1; i < n; ++i) {
    int x, y, w;
    scanf("%d%d%d", &x, &y, &w);
    edge[x].emplace_back(y, w);
    edge[y].emplace_back(x, w);
  }
  dfs(1, 0);
  dfs2(1, 0, 0);
  dfs3(1, 0);
  while (q--) {
    int x, y, p;
    scanf("%d%d", &x, &y);
    p = lca(x, y);
    long long vs = sump[x] + sump[y] - sump[par[p][0]] - sump[p];
    long long es = sums[x] + sums[y] - (sums[p] << 1);
    long long ls = len[x] + len[y] - (len[p] << 1);
    printf("%lld\n", vs - es - ls);
  }
  return 0;
}

```