Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employee ai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi. 

Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.

Input

The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 106)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application qai > qbi.

Output

Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.

Examples

Input

4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5


Output

11


Input

3
1 2 3
2
3 1 2
3 1 3


Output

-1

Note

In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, a, b, c;
  cin >> n;
  for (int i = 0; i <= n; ++i) cin >> m;
  vector<vector<pair<int, int> > > E(n);
  for (int i = 0; i < m; ++i) {
    cin >> a >> b >> c;
    --a, --b;
    auto it = E[b].begin();
    for (; it != E[b].end(); ++it)
      if (it->first == a) {
        break;
      }
    if (it == E[b].end())
      E[b].push_back(make_pair(a, c));
    else
      it->second = min(it->second, c);
  }
  const int inf = 1 << 30;
  vector<int> used(n, 0);
  int ans = 0;
  for (int it = 1; it < n; ++it) {
    int v = -1;
    bool find = false;
    int min_c = inf;
    for (int u = 0; u < n && !find; ++u)
      if (!used[u])
        for (auto i = E[u].begin(); i != E[u].end() && !find; ++i)
          if (i->second < min_c) {
            min_c = i->second;
            v = u;
          }
    if (min_c == inf) {
      cout << -1 << endl;
      return 0;
    }
    used[v] = true;
    ans += min_c;
  }
  cout << ans << endl;
  return 0;
}
