#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5;
long long a[maxn + 5], e[maxn + 5];
int fa[maxn + 5][20], dep[maxn + 5];
long long s[maxn + 5], ss[maxn + 5], se[maxn + 5];
long long f[maxn + 5], dp[maxn + 5];
vector<pair<int, int> > g[maxn + 5];
int n, m;
void dfs(int k, int last) {
  dep[k] = dep[last] + 1;
  fa[k][0] = last;
  f[k] = a[k];
  for (auto x : g[k])
    if (x.first != last) {
      e[x.first] = x.second;
      dfs(x.first, k);
      if (f[x.first] - 2 * x.second > 0) f[k] += f[x.first] - 2 * x.second;
    }
}
void dfs2(int k, int last) {
  s[k] = s[last] + f[k];
  se[k] = se[last] + e[k];
  ss[k] = ss[last];
  if (f[k] - 2 * e[k] > 0) ss[k] += f[k] - 2 * e[k];
  for (auto x : g[k])
    if (x.first != last) dfs2(x.first, k);
}
void dfs1(int k, int last) {
  long long now = 0;
  for (int i = 0; i < g[k].size(); ++i) {
    pair<int, int> x = g[k][i];
    if (x.first != last) {
      dp[x.first] += now;
      if (f[x.first] - 2 * x.second > 0) now += f[x.first] - 2 * x.second;
    }
  }
  now = 0;
  for (int i = g[k].size() - 1; i >= 0; --i) {
    pair<int, int> x = g[k][i];
    if (x.first != last) {
      dp[x.first] += now;
      if (f[x.first] - 2 * x.second > 0) now += f[x.first] - 2 * x.second;
    }
  }
  for (int i = 0; i < g[k].size(); ++i) {
    pair<int, int> x = g[k][i];
    if (x.first != last) {
      dp[x.first] += dp[k] - 2 * x.second;
      dp[x.first] = max(0LL, dp[x.first]);
      dp[x.first] += a[x.first];
    }
  }
  for (auto x : g[k])
    if (x.first != last) dfs1(x.first, k);
}
int lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  int x = dep[u] - dep[v];
  for (int i = 0; i < 20; ++i)
    if ((1 << i) & x) u = fa[u][i];
  for (int i = 19; i >= 0; --i)
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  if (u == v)
    return u;
  else
    return fa[u][0];
}
long long query(int u, int v) {
  long long ans = 0;
  int w = lca(u, v);
  ans += s[u] + s[v] - 2 * s[w];
  ans -= ss[u] + ss[v] - 2 * ss[w];
  ans += dp[w] + f[w] - a[w];
  ans -= se[u] + se[v] - 2 * se[w];
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    g[u].push_back(make_pair(v, w));
    g[v].push_back(make_pair(u, w));
  }
  dfs(1, 0);
  dfs2(1, 0);
  dp[1] = a[1];
  dfs1(1, 0);
  for (int j = 1; j < 20; ++j)
    for (int i = 1; i <= n; ++i) fa[i][j] = fa[fa[i][j - 1]][j - 1];
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%lld\n", query(l, r));
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 = 3e5;
long long a[maxn + 5], e[maxn + 5];
int fa[maxn + 5][20], dep[maxn + 5];
long long s[maxn + 5], ss[maxn + 5], se[maxn + 5];
long long f[maxn + 5], dp[maxn + 5];
vector<pair<int, int> > g[maxn + 5];
int n, m;
void dfs(int k, int last) {
  dep[k] = dep[last] + 1;
  fa[k][0] = last;
  f[k] = a[k];
  for (auto x : g[k])
    if (x.first != last) {
      e[x.first] = x.second;
      dfs(x.first, k);
      if (f[x.first] - 2 * x.second > 0) f[k] += f[x.first] - 2 * x.second;
    }
}
void dfs2(int k, int last) {
  s[k] = s[last] + f[k];
  se[k] = se[last] + e[k];
  ss[k] = ss[last];
  if (f[k] - 2 * e[k] > 0) ss[k] += f[k] - 2 * e[k];
  for (auto x : g[k])
    if (x.first != last) dfs2(x.first, k);
}
void dfs1(int k, int last) {
  long long now = 0;
  for (int i = 0; i < g[k].size(); ++i) {
    pair<int, int> x = g[k][i];
    if (x.first != last) {
      dp[x.first] += now;
      if (f[x.first] - 2 * x.second > 0) now += f[x.first] - 2 * x.second;
    }
  }
  now = 0;
  for (int i = g[k].size() - 1; i >= 0; --i) {
    pair<int, int> x = g[k][i];
    if (x.first != last) {
      dp[x.first] += now;
      if (f[x.first] - 2 * x.second > 0) now += f[x.first] - 2 * x.second;
    }
  }
  for (int i = 0; i < g[k].size(); ++i) {
    pair<int, int> x = g[k][i];
    if (x.first != last) {
      dp[x.first] += dp[k] - 2 * x.second;
      dp[x.first] = max(0LL, dp[x.first]);
      dp[x.first] += a[x.first];
    }
  }
  for (auto x : g[k])
    if (x.first != last) dfs1(x.first, k);
}
int lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  int x = dep[u] - dep[v];
  for (int i = 0; i < 20; ++i)
    if ((1 << i) & x) u = fa[u][i];
  for (int i = 19; i >= 0; --i)
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  if (u == v)
    return u;
  else
    return fa[u][0];
}
long long query(int u, int v) {
  long long ans = 0;
  int w = lca(u, v);
  ans += s[u] + s[v] - 2 * s[w];
  ans -= ss[u] + ss[v] - 2 * ss[w];
  ans += dp[w] + f[w] - a[w];
  ans -= se[u] + se[v] - 2 * se[w];
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    g[u].push_back(make_pair(v, w));
    g[v].push_back(make_pair(u, w));
  }
  dfs(1, 0);
  dfs2(1, 0);
  dp[1] = a[1];
  dfs1(1, 0);
  for (int j = 1; j < 20; ++j)
    for (int i = 1; i <= n; ++i) fa[i][j] = fa[fa[i][j - 1]][j - 1];
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%lld\n", query(l, r));
  }
  return 0;
}

```