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;
bool mycomp(pair<long int, long int> a, pair<long int, long int> b) {
  if (a.first != b.first) return a.first < b.first;
  return a.second < b.second;
}
int main() {
  long int n;
  cin >> n;
  long int a[n + 2];
  for (long int i = 1; i <= n; i++) cin >> a[i];
  long int m;
  cin >> m;
  set<long int> s;
  vector<pair<long int, long int> > v;
  for (long int i = 0; i < m; i++) {
    long int a, b, c;
    cin >> a >> b >> c;
    s.insert(b);
    v.push_back(make_pair(b, c));
  }
  if (s.size() == n - 1) {
    sort(v.begin(), v.end(), mycomp);
    long long int sum = 0;
    for (long int i = 0; i < v.size(); i++) {
      sum += v[i].second;
      long int ff = v[i].first;
      while (v[i + 1].first == ff && i < v.size()) i++;
    }
    cout << sum << endl;
  } else {
    cout << "-1\n";
  }
}
