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;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n;
  cin >> n;
  vector<long long int> q(n + 1);
  for (long long int i = 1; i <= n; i++) cin >> q[i];
  long long int m;
  cin >> m;
  vector<pair<long long int, long long int>> adj[n + 1];
  for (long long int i = 1; i <= m; i++) {
    long long int a, b, c;
    cin >> a >> b >> c;
    adj[b].push_back({c, a});
  }
  long long int cnt = 0;
  for (long long int i = 1; i <= n; i++) {
    sort(adj[i].begin(), adj[i].end());
    if (adj[i].size() > 0) cnt++;
  }
  if (cnt == n - 1) {
    long long int ans = 0;
    for (long long int i = 1; i <= n; i++) {
      if (adj[i].size() > 0) ans += adj[i][0].first;
    }
    cout << ans;
  } else {
    cout << -1;
  }
}
