#include <bits/stdc++.h>
using namespace std;
const long long N = 3e5 + 5;
long long n, m, a[N], pnt[N * 2], nxt[N * 2], val[N * 2], head[N], E;
long long down[N], up[N], far[N][20], dep[N], ad[N], dl[N];
long long infu[N], lth[N], sumv[N], sumd[N];
void add(long long u, long long v, long long w) {
  pnt[++E] = v;
  nxt[E] = head[u];
  val[E] = w;
  head[u] = E;
}
void sol_down(long long u, long long fa) {
  down[u] = a[u];
  for (long long i = head[u]; i; i = nxt[i]) {
    long long v = pnt[i];
    if (v == fa) continue;
    sol_down(v, u);
    if (down[v] - val[i] * 2 > 0) {
      down[u] += down[v] - val[i] * 2;
      infu[v] = down[v] - val[i] * 2;
    }
  }
}
void sol_up(long long u, long long fa) {
  sumd[u] = sumd[fa] + down[u];
  sumv[u] = sumv[fa] + a[u];
  infu[u] += infu[fa];
  for (long long i = head[u]; i; i = nxt[i]) {
    long long v = pnt[i];
    if (v == fa) continue;
    up[v] = max(up[u] + down[u] - infu[v] - val[i] * 2, 0ll);
    far[v][0] = u;
    dep[v] = dep[u] + 1;
    lth[v] = lth[u] + val[i];
    sol_up(v, u);
  }
}
long long LCA(long long u, long long v) {
  if (dep[u] < dep[v]) swap(u, v);
  long long dlt = dep[u] - dep[v];
  for (long long i = 19; i >= 0; --i)
    if (dlt & (1 << i)) u = far[u][i];
  if (u == v) return u;
  for (long long i = 19; i >= 0; --i)
    if (far[u][i] ^ far[v][i]) u = far[u][i], v = far[v][i];
  return far[u][0];
}
signed main() {
  cin >> n >> m;
  for (long long i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  for (long long u, v, w, i = 2; i <= n; ++i) {
    scanf("%lld%lld%lld", &u, &v, &w);
    add(u, v, w);
    add(v, u, w);
  }
  sol_down(1, 0);
  sol_up(1, 0);
  for (long long i = 1; i < 20; ++i)
    for (long long u = 1; u <= n; ++u) far[u][i] = far[far[u][i - 1]][i - 1];
  long long u, v, lca, res;
  while (m-- > 0) {
    scanf("%lld%lld", &u, &v);
    if (u == v)
      res = down[u] + up[u];
    else {
      lca = LCA(u, v);
      res = 0;
      res += up[lca];
      res += sumd[u] - sumd[lca];
      res += sumd[v] - sumd[lca];
      res += down[lca];
      res -= infu[u] - infu[lca];
      res -= infu[v] - infu[lca];
      res -= lth[u] + lth[v] - lth[lca] * 2;
    }
    printf("%lld\n", res);
  }
  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>
using namespace std;
const long long N = 3e5 + 5;
long long n, m, a[N], pnt[N * 2], nxt[N * 2], val[N * 2], head[N], E;
long long down[N], up[N], far[N][20], dep[N], ad[N], dl[N];
long long infu[N], lth[N], sumv[N], sumd[N];
void add(long long u, long long v, long long w) {
  pnt[++E] = v;
  nxt[E] = head[u];
  val[E] = w;
  head[u] = E;
}
void sol_down(long long u, long long fa) {
  down[u] = a[u];
  for (long long i = head[u]; i; i = nxt[i]) {
    long long v = pnt[i];
    if (v == fa) continue;
    sol_down(v, u);
    if (down[v] - val[i] * 2 > 0) {
      down[u] += down[v] - val[i] * 2;
      infu[v] = down[v] - val[i] * 2;
    }
  }
}
void sol_up(long long u, long long fa) {
  sumd[u] = sumd[fa] + down[u];
  sumv[u] = sumv[fa] + a[u];
  infu[u] += infu[fa];
  for (long long i = head[u]; i; i = nxt[i]) {
    long long v = pnt[i];
    if (v == fa) continue;
    up[v] = max(up[u] + down[u] - infu[v] - val[i] * 2, 0ll);
    far[v][0] = u;
    dep[v] = dep[u] + 1;
    lth[v] = lth[u] + val[i];
    sol_up(v, u);
  }
}
long long LCA(long long u, long long v) {
  if (dep[u] < dep[v]) swap(u, v);
  long long dlt = dep[u] - dep[v];
  for (long long i = 19; i >= 0; --i)
    if (dlt & (1 << i)) u = far[u][i];
  if (u == v) return u;
  for (long long i = 19; i >= 0; --i)
    if (far[u][i] ^ far[v][i]) u = far[u][i], v = far[v][i];
  return far[u][0];
}
signed main() {
  cin >> n >> m;
  for (long long i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  for (long long u, v, w, i = 2; i <= n; ++i) {
    scanf("%lld%lld%lld", &u, &v, &w);
    add(u, v, w);
    add(v, u, w);
  }
  sol_down(1, 0);
  sol_up(1, 0);
  for (long long i = 1; i < 20; ++i)
    for (long long u = 1; u <= n; ++u) far[u][i] = far[far[u][i - 1]][i - 1];
  long long u, v, lca, res;
  while (m-- > 0) {
    scanf("%lld%lld", &u, &v);
    if (u == v)
      res = down[u] + up[u];
    else {
      lca = LCA(u, v);
      res = 0;
      res += up[lca];
      res += sumd[u] - sumd[lca];
      res += sumd[v] - sumd[lca];
      res += down[lca];
      res -= infu[u] - infu[lca];
      res -= infu[v] - infu[lca];
      res -= lth[u] + lth[v] - lth[lca] * 2;
    }
    printf("%lld\n", res);
  }
  return 0;
}

```