#include <bits/stdc++.h>
using namespace std;
int n, q;
int a[300010];
struct Edge {
  int v, c, nxt;
} e[600010];
int tot;
int first[300010];
void build(int u, int v, int c) {
  e[++tot] = (Edge){v, c, first[u]};
  first[u] = tot;
  return;
}
int dep[300010];
long long f[300010];
long long g[300010];
long long sum[300010];
const int M = 20;
int fa[300010][M];
void dfs1(int u, int lst) {
  dep[u] = dep[lst] + 1;
  fa[u][0] = lst;
  for (int i = 1; i < M; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  f[u] = 0;
  for (int i = first[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == lst) continue;
    dfs1(v, u);
    f[u] += max(0ll, f[v] + a[v] - 2ll * e[i].c);
  }
  return;
}
void dfs2(int u, int lst) {
  sum[u] += a[u] + f[u];
  for (int i = first[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == lst) continue;
    long long val = max(0ll, f[v] + a[v] - 2ll * e[i].c);
    g[v] = max(0ll, f[u] + g[u] - val + a[u] - 2ll * e[i].c);
    sum[v] += sum[u] - val - e[i].c;
    dfs2(v, u);
  }
  return;
}
int getlca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  int need = dep[u] - dep[v], now = 0;
  while (need) {
    if (need & 1) u = fa[u][now];
    need >>= 1;
    now++;
  }
  if (u == v) return u;
  for (int i = M - 1; i >= 0; i--)
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  return fa[u][0];
}
long long solve(int u, int v) {
  int lca = getlca(u, v);
  return sum[u] + sum[v] - 2 * sum[lca] + f[lca] + g[lca] + a[lca];
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    int u, v, c;
    scanf("%d %d %d", &u, &v, &c);
    build(u, v, c);
    build(v, u, c);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  while (q--) {
    int u, v;
    scanf("%d %d", &u, &v);
    printf("%lld\n", solve(u, v));
  }
  return 0;
}

### Prompt

Generate 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;
int n, q;
int a[300010];
struct Edge {
  int v, c, nxt;
} e[600010];
int tot;
int first[300010];
void build(int u, int v, int c) {
  e[++tot] = (Edge){v, c, first[u]};
  first[u] = tot;
  return;
}
int dep[300010];
long long f[300010];
long long g[300010];
long long sum[300010];
const int M = 20;
int fa[300010][M];
void dfs1(int u, int lst) {
  dep[u] = dep[lst] + 1;
  fa[u][0] = lst;
  for (int i = 1; i < M; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  f[u] = 0;
  for (int i = first[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == lst) continue;
    dfs1(v, u);
    f[u] += max(0ll, f[v] + a[v] - 2ll * e[i].c);
  }
  return;
}
void dfs2(int u, int lst) {
  sum[u] += a[u] + f[u];
  for (int i = first[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == lst) continue;
    long long val = max(0ll, f[v] + a[v] - 2ll * e[i].c);
    g[v] = max(0ll, f[u] + g[u] - val + a[u] - 2ll * e[i].c);
    sum[v] += sum[u] - val - e[i].c;
    dfs2(v, u);
  }
  return;
}
int getlca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  int need = dep[u] - dep[v], now = 0;
  while (need) {
    if (need & 1) u = fa[u][now];
    need >>= 1;
    now++;
  }
  if (u == v) return u;
  for (int i = M - 1; i >= 0; i--)
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  return fa[u][0];
}
long long solve(int u, int v) {
  int lca = getlca(u, v);
  return sum[u] + sum[v] - 2 * sum[lca] + f[lca] + g[lca] + a[lca];
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    int u, v, c;
    scanf("%d %d %d", &u, &v, &c);
    build(u, v, c);
    build(v, u, c);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  while (q--) {
    int u, v;
    scanf("%d %d", &u, &v);
    printf("%lld\n", solve(u, v));
  }
  return 0;
}

```