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.
import java.util.*;

public class Main {
    static final int INF = 2000000;
    public static void main(String[] args){
        Scanner s = new Scanner(System.in);
        final int n = s.nextInt();
        for (int i = 0; i < n; ++i){
            s.nextInt();
        }
        final int m = s.nextInt();
        int [] cost = new int[n + 1];
        Arrays.fill(cost, INF);
        for (int i = 0; i < m; ++i){
            final int a = s.nextInt();
            final int b = s.nextInt();
            final int c = s.nextInt();
            cost[b] = Math.min(cost[b], c);
        }
        int c = 0;
        int res = 0;
        for (int i = 1; i <= n; ++i){
            if (cost[i] == INF){
                c++;
            }else{
                res += cost[i];
            }
        }
        if (c > 1){
            System.out.println(-1);
        }else{
            System.out.println(res);
        }
    }
}
