#include <bits/stdc++.h>
using namespace std;
const int maxn = 300007;
int pre[maxn][20], n, a[maxn], head[maxn], q, dep[maxn];
struct EDGE {
  int to, nxt, w;
} edge[maxn << 1];
void addedge(int u, int v, int w) {
  static int sz = 0;
  edge[++sz].nxt = head[u];
  edge[sz].to = v;
  edge[sz].w = w;
  head[u] = sz;
}
long long f[maxn], g[maxn], sum[maxn][20], up[maxn], mi[maxn][20], from[maxn];
void dfs(int u) {
  int to;
  f[u] = a[u];
  dep[u] = dep[pre[u][0]] + 1;
  mi[u][0] = a[pre[u][0]] - from[u];
  for (int i = 1; i < 20; ++i) {
    pre[u][i] = pre[pre[u][i - 1]][i - 1];
    mi[u][i] = mi[u][i - 1] + mi[pre[u][i - 1]][i - 1];
  }
  for (int i = head[u]; i; i = edge[i].nxt) {
    to = edge[i].to;
    if (to == pre[u][0]) continue;
    from[to] = edge[i].w;
    pre[to][0] = u;
    dfs(to);
    if (f[to] - 2 * edge[i].w > 0) f[u] += f[to] - 2 * edge[i].w;
  }
  for (int i = head[u]; i; i = edge[i].nxt) {
    to = edge[i].to;
    if (to == pre[u][0]) continue;
    if (f[to] - 2 * edge[i].w > 0)
      g[to] = f[u] - a[u] - f[to] + 2 * edge[i].w;
    else
      g[to] = f[u] - a[u];
  }
}
void dfs2(int u) {
  sum[u][0] = g[u];
  up[u] = max(0ll, up[pre[u][0]] + a[pre[u][0]] + g[u] - 2 * from[u]);
  for (int i = 1; i < 20; ++i)
    sum[u][i] = sum[u][i - 1] + sum[pre[u][i - 1]][i - 1];
  int to;
  for (int i = head[u]; i; i = edge[i].nxt) {
    to = edge[i].to;
    if (to == pre[u][0]) continue;
    dfs2(to);
  }
}
pair<int, long long> getlca(int u, int v) {
  long long SUM = 0;
  if (dep[u] < dep[v]) swap(u, v);
  int p1 = u, p2 = v;
  for (int i = 19; i >= 0; --i) {
    if (dep[pre[u][i]] >= dep[v]) {
      SUM += sum[u][i] + mi[u][i];
      u = pre[u][i];
    }
  }
  if (u == v) return make_pair(u, SUM + f[p1]);
  for (int i = 19; i >= 0; --i) {
    if (pre[u][i] != pre[v][i]) {
      SUM += sum[u][i] + mi[u][i];
      SUM += sum[v][i] + mi[v][i];
      u = pre[u][i];
      v = pre[v][i];
    }
  }
  SUM += sum[u][0] + mi[u][0];
  SUM += sum[v][0] + mi[v][0];
  SUM -= f[pre[u][0]] - a[pre[u][0]];
  return make_pair(pre[u][0], SUM + f[p1] + f[p2] - a[pre[u][0]]);
}
long long solve(int u, int v) {
  pair<int, long long> p = getlca(u, v);
  return p.second + up[p.first];
}
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) {
    static int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    addedge(u, v, w);
    addedge(v, u, w);
  }
  dfs(1);
  dfs2(1);
  while (q--) {
    static 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;
const int maxn = 300007;
int pre[maxn][20], n, a[maxn], head[maxn], q, dep[maxn];
struct EDGE {
  int to, nxt, w;
} edge[maxn << 1];
void addedge(int u, int v, int w) {
  static int sz = 0;
  edge[++sz].nxt = head[u];
  edge[sz].to = v;
  edge[sz].w = w;
  head[u] = sz;
}
long long f[maxn], g[maxn], sum[maxn][20], up[maxn], mi[maxn][20], from[maxn];
void dfs(int u) {
  int to;
  f[u] = a[u];
  dep[u] = dep[pre[u][0]] + 1;
  mi[u][0] = a[pre[u][0]] - from[u];
  for (int i = 1; i < 20; ++i) {
    pre[u][i] = pre[pre[u][i - 1]][i - 1];
    mi[u][i] = mi[u][i - 1] + mi[pre[u][i - 1]][i - 1];
  }
  for (int i = head[u]; i; i = edge[i].nxt) {
    to = edge[i].to;
    if (to == pre[u][0]) continue;
    from[to] = edge[i].w;
    pre[to][0] = u;
    dfs(to);
    if (f[to] - 2 * edge[i].w > 0) f[u] += f[to] - 2 * edge[i].w;
  }
  for (int i = head[u]; i; i = edge[i].nxt) {
    to = edge[i].to;
    if (to == pre[u][0]) continue;
    if (f[to] - 2 * edge[i].w > 0)
      g[to] = f[u] - a[u] - f[to] + 2 * edge[i].w;
    else
      g[to] = f[u] - a[u];
  }
}
void dfs2(int u) {
  sum[u][0] = g[u];
  up[u] = max(0ll, up[pre[u][0]] + a[pre[u][0]] + g[u] - 2 * from[u]);
  for (int i = 1; i < 20; ++i)
    sum[u][i] = sum[u][i - 1] + sum[pre[u][i - 1]][i - 1];
  int to;
  for (int i = head[u]; i; i = edge[i].nxt) {
    to = edge[i].to;
    if (to == pre[u][0]) continue;
    dfs2(to);
  }
}
pair<int, long long> getlca(int u, int v) {
  long long SUM = 0;
  if (dep[u] < dep[v]) swap(u, v);
  int p1 = u, p2 = v;
  for (int i = 19; i >= 0; --i) {
    if (dep[pre[u][i]] >= dep[v]) {
      SUM += sum[u][i] + mi[u][i];
      u = pre[u][i];
    }
  }
  if (u == v) return make_pair(u, SUM + f[p1]);
  for (int i = 19; i >= 0; --i) {
    if (pre[u][i] != pre[v][i]) {
      SUM += sum[u][i] + mi[u][i];
      SUM += sum[v][i] + mi[v][i];
      u = pre[u][i];
      v = pre[v][i];
    }
  }
  SUM += sum[u][0] + mi[u][0];
  SUM += sum[v][0] + mi[v][0];
  SUM -= f[pre[u][0]] - a[pre[u][0]];
  return make_pair(pre[u][0], SUM + f[p1] + f[p2] - a[pre[u][0]]);
}
long long solve(int u, int v) {
  pair<int, long long> p = getlca(u, v);
  return p.second + up[p.first];
}
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) {
    static int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    addedge(u, v, w);
    addedge(v, u, w);
  }
  dfs(1);
  dfs2(1);
  while (q--) {
    static int u, v;
    scanf("%d%d", &u, &v);
    printf("%lld\n", solve(u, v));
  }
  return 0;
}

```