#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
struct edge {
  int v, w, next;
} e[300005 << 1];
int head[300005], k;
inline void add(int u, int v, int w) {
  e[k] = (edge){v, w, head[u]};
  head[u] = k++;
}
int n, q;
long long f[300005], a[300005];
inline void dp(int u, int fa) {
  f[u] = a[u];
  for (int i = head[u]; ~i; i = e[i].next) {
    int v = e[i].v;
    if (v == fa) continue;
    dp(v, u);
    if (f[v] - e[i].w - e[i ^ 1].w > 0) f[u] += f[v] - e[i].w - e[i ^ 1].w;
  }
}
int p[300005][30];
long long F1[300005][30], F2[300005][30], g[300005];
int dep[300005];
inline void dfs(int u, int fa) {
  dep[u] = dep[p[u][0] = fa] + 1;
  for (int i = 1; i <= 25; ++i) {
    p[u][i] = p[p[u][i - 1]][i - 1];
    F1[u][i] = F1[u][i - 1] + F1[p[u][i - 1]][i - 1] - f[p[u][i - 1]];
    F2[u][i] = F2[u][i - 1] + F2[p[u][i - 1]][i - 1] - f[p[u][i - 1]];
  }
  for (int i = head[u]; ~i; i = e[i].next) {
    int v = e[i].v;
    if (v == fa) continue;
    F1[v][0] = f[v] + f[u] - e[i ^ 1].w;
    if (f[v] - e[i].w - e[i ^ 1].w > 0) F1[v][0] -= f[v] - e[i].w - e[i ^ 1].w;
    F2[v][0] = f[v] + f[u] - e[i].w;
    if (f[v] - e[i].w - e[i ^ 1].w > 0) F2[v][0] -= f[v] - e[i].w - e[i ^ 1].w;
    g[v] = max(g[v], g[u] + F1[v][0] - e[i].w - f[v]);
    dfs(v, u);
  }
}
inline long long Query(int u, int v) {
  long long ans = 0;
  int a = u, b = v;
  if (dep[u] > dep[v]) {
    for (int i = 25; i >= 0; --i)
      if (dep[p[u][i]] >= dep[v]) {
        ans += F1[u][i] - f[u];
        u = p[u][i];
      }
    if (u == v) return ans + f[a] + g[u];
  } else {
    for (int i = 25; i >= 0; --i)
      if (dep[p[v][i]] >= dep[u]) {
        ans += F2[v][i] - f[v];
        v = p[v][i];
      }
    if (u == v) return ans + f[b] + g[u];
  }
  for (int i = 25; i >= 0; --i)
    if (p[u][i] != p[v][i]) {
      ans += F1[u][i] - f[u];
      ans += F2[v][i] - f[v];
      u = p[u][i];
      v = p[v][i];
    }
  return ans + F1[u][0] - f[u] + F2[v][0] - f[v] - f[p[u][0]] + f[a] + f[b] +
         g[p[u][0]];
}
int main() {
  n = read(), q = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  memset(head, -1, sizeof(int) * (n + 5));
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read();
    long long w = read();
    add(u, v, w);
    add(v, u, w);
  }
  dp(1, 0);
  dfs(1, 0);
  while (q--) {
    int u = read(), v = read();
    printf("%lld\n", Query(u, v));
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
struct edge {
  int v, w, next;
} e[300005 << 1];
int head[300005], k;
inline void add(int u, int v, int w) {
  e[k] = (edge){v, w, head[u]};
  head[u] = k++;
}
int n, q;
long long f[300005], a[300005];
inline void dp(int u, int fa) {
  f[u] = a[u];
  for (int i = head[u]; ~i; i = e[i].next) {
    int v = e[i].v;
    if (v == fa) continue;
    dp(v, u);
    if (f[v] - e[i].w - e[i ^ 1].w > 0) f[u] += f[v] - e[i].w - e[i ^ 1].w;
  }
}
int p[300005][30];
long long F1[300005][30], F2[300005][30], g[300005];
int dep[300005];
inline void dfs(int u, int fa) {
  dep[u] = dep[p[u][0] = fa] + 1;
  for (int i = 1; i <= 25; ++i) {
    p[u][i] = p[p[u][i - 1]][i - 1];
    F1[u][i] = F1[u][i - 1] + F1[p[u][i - 1]][i - 1] - f[p[u][i - 1]];
    F2[u][i] = F2[u][i - 1] + F2[p[u][i - 1]][i - 1] - f[p[u][i - 1]];
  }
  for (int i = head[u]; ~i; i = e[i].next) {
    int v = e[i].v;
    if (v == fa) continue;
    F1[v][0] = f[v] + f[u] - e[i ^ 1].w;
    if (f[v] - e[i].w - e[i ^ 1].w > 0) F1[v][0] -= f[v] - e[i].w - e[i ^ 1].w;
    F2[v][0] = f[v] + f[u] - e[i].w;
    if (f[v] - e[i].w - e[i ^ 1].w > 0) F2[v][0] -= f[v] - e[i].w - e[i ^ 1].w;
    g[v] = max(g[v], g[u] + F1[v][0] - e[i].w - f[v]);
    dfs(v, u);
  }
}
inline long long Query(int u, int v) {
  long long ans = 0;
  int a = u, b = v;
  if (dep[u] > dep[v]) {
    for (int i = 25; i >= 0; --i)
      if (dep[p[u][i]] >= dep[v]) {
        ans += F1[u][i] - f[u];
        u = p[u][i];
      }
    if (u == v) return ans + f[a] + g[u];
  } else {
    for (int i = 25; i >= 0; --i)
      if (dep[p[v][i]] >= dep[u]) {
        ans += F2[v][i] - f[v];
        v = p[v][i];
      }
    if (u == v) return ans + f[b] + g[u];
  }
  for (int i = 25; i >= 0; --i)
    if (p[u][i] != p[v][i]) {
      ans += F1[u][i] - f[u];
      ans += F2[v][i] - f[v];
      u = p[u][i];
      v = p[v][i];
    }
  return ans + F1[u][0] - f[u] + F2[v][0] - f[v] - f[p[u][0]] + f[a] + f[b] +
         g[p[u][0]];
}
int main() {
  n = read(), q = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  memset(head, -1, sizeof(int) * (n + 5));
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read();
    long long w = read();
    add(u, v, w);
    add(v, u, w);
  }
  dp(1, 0);
  dfs(1, 0);
  while (q--) {
    int u = read(), v = read();
    printf("%lld\n", Query(u, v));
  }
  return 0;
}

```