#include <bits/stdc++.h>
using namespace std;
const long long N = 3e5 + 9;
long long n, m, head[N], tot;
long long a[N];
struct pp {
  long long nxt, to;
  long long val;
} g[N << 1];
void add(long long u, long long v, long long w) {
  g[++tot].nxt = head[u], g[tot].to = v, g[tot].val = w, head[u] = tot;
}
long long f[N][22], dep[N];
long long dp[N], up[N], w[N][22], Dp[N][22];
void dfs(long long u, long long fa) {
  f[u][0] = fa, dep[u] = dep[fa] + 1;
  long long cnt = 0;
  dp[u] = 0;
  for (long long i = head[u]; i != -1; i = g[i].nxt) {
    long long v = g[i].to;
    if (v == fa) continue;
    dfs(v, u);
    cnt++;
    dp[u] += max(dp[v] - 2ll * g[i].val, 0ll);
  }
  if (!cnt) {
    dp[u] = a[u];
    return;
  }
  dp[u] += a[u];
  return;
}
void dfs2(long long u, long long fa) {
  Dp[u][0] = dp[u];
  for (long long i = head[u]; i != -1; i = g[i].nxt) {
    long long v = g[i].to;
    if (v == fa) continue;
    up[v] =
        max(up[u] + dp[u] - max(dp[v] - 2 * g[i].val, 0ll) - 2 * g[i].val, 0ll);
    w[v][0] = max(dp[v] - 2 * g[i].val, 0ll) + g[i].val;
    dfs2(v, u);
  }
}
long long lca(long long u, long long v) {
  if (u == v) return dp[u] + up[u];
  long long ans = 0;
  if (dep[u] < dep[v]) swap(u, v);
  for (long long i = 21; i >= 0; i--)
    if (dep[f[u][i]] >= dep[v]) {
      ans += Dp[u][i];
      ans -= w[u][i];
      u = f[u][i];
    }
  if (u == v) {
    ans += up[u];
    ans += dp[u];
    return ans;
  }
  for (long long i = 21; i >= 0; i--) {
    if (f[u][i] != f[v][i]) {
      ans += Dp[u][i];
      ans -= w[u][i];
      ans += Dp[v][i];
      ans -= w[v][i];
      u = f[u][i], v = f[v][i];
    }
  }
  ans += dp[u];
  ans += dp[v];
  ans -= w[u][0], ans -= w[v][0];
  ans += Dp[f[u][0]][0];
  ans += up[f[u][0]];
  return ans;
}
signed main() {
  memset(head, -1, sizeof(head));
  tot = 0;
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1; i < n; i++) {
    long long u, v, w;
    scanf("%lld%lld%lld", &u, &v, &w);
    add(u, v, w);
    add(v, u, w);
  }
  dep[0] = 0;
  dfs(1, 0);
  dfs2(1, 0);
  for (long long j = 1; j <= 21; j++)
    for (long long i = 1; i <= n; i++)
      f[i][j] = f[f[i][j - 1]][j - 1],
      w[i][j] = w[i][j - 1] + w[f[i][j - 1]][j - 1];
  for (long long j = 1; j <= 21; j++)
    for (long long i = 1; i <= n; i++)
      Dp[i][j] = Dp[i][j - 1] + Dp[f[i][j - 1]][j - 1];
  while (m--) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    printf("%lld\n", lca(u, v));
  }
  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 long long N = 3e5 + 9;
long long n, m, head[N], tot;
long long a[N];
struct pp {
  long long nxt, to;
  long long val;
} g[N << 1];
void add(long long u, long long v, long long w) {
  g[++tot].nxt = head[u], g[tot].to = v, g[tot].val = w, head[u] = tot;
}
long long f[N][22], dep[N];
long long dp[N], up[N], w[N][22], Dp[N][22];
void dfs(long long u, long long fa) {
  f[u][0] = fa, dep[u] = dep[fa] + 1;
  long long cnt = 0;
  dp[u] = 0;
  for (long long i = head[u]; i != -1; i = g[i].nxt) {
    long long v = g[i].to;
    if (v == fa) continue;
    dfs(v, u);
    cnt++;
    dp[u] += max(dp[v] - 2ll * g[i].val, 0ll);
  }
  if (!cnt) {
    dp[u] = a[u];
    return;
  }
  dp[u] += a[u];
  return;
}
void dfs2(long long u, long long fa) {
  Dp[u][0] = dp[u];
  for (long long i = head[u]; i != -1; i = g[i].nxt) {
    long long v = g[i].to;
    if (v == fa) continue;
    up[v] =
        max(up[u] + dp[u] - max(dp[v] - 2 * g[i].val, 0ll) - 2 * g[i].val, 0ll);
    w[v][0] = max(dp[v] - 2 * g[i].val, 0ll) + g[i].val;
    dfs2(v, u);
  }
}
long long lca(long long u, long long v) {
  if (u == v) return dp[u] + up[u];
  long long ans = 0;
  if (dep[u] < dep[v]) swap(u, v);
  for (long long i = 21; i >= 0; i--)
    if (dep[f[u][i]] >= dep[v]) {
      ans += Dp[u][i];
      ans -= w[u][i];
      u = f[u][i];
    }
  if (u == v) {
    ans += up[u];
    ans += dp[u];
    return ans;
  }
  for (long long i = 21; i >= 0; i--) {
    if (f[u][i] != f[v][i]) {
      ans += Dp[u][i];
      ans -= w[u][i];
      ans += Dp[v][i];
      ans -= w[v][i];
      u = f[u][i], v = f[v][i];
    }
  }
  ans += dp[u];
  ans += dp[v];
  ans -= w[u][0], ans -= w[v][0];
  ans += Dp[f[u][0]][0];
  ans += up[f[u][0]];
  return ans;
}
signed main() {
  memset(head, -1, sizeof(head));
  tot = 0;
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1; i < n; i++) {
    long long u, v, w;
    scanf("%lld%lld%lld", &u, &v, &w);
    add(u, v, w);
    add(v, u, w);
  }
  dep[0] = 0;
  dfs(1, 0);
  dfs2(1, 0);
  for (long long j = 1; j <= 21; j++)
    for (long long i = 1; i <= n; i++)
      f[i][j] = f[f[i][j - 1]][j - 1],
      w[i][j] = w[i][j - 1] + w[f[i][j - 1]][j - 1];
  for (long long j = 1; j <= 21; j++)
    for (long long i = 1; i <= n; i++)
      Dp[i][j] = Dp[i][j - 1] + Dp[f[i][j - 1]][j - 1];
  while (m--) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    printf("%lld\n", lca(u, v));
  }
  return 0;
}

```