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 Emps {

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int[] q = new int[n];
        for (int i = 0; i < n; i++) {
            q[i] = s.nextInt();
        }
        int m = s.nextInt();
        HashMap<Integer, Integer> sal = new HashMap<Integer, Integer>();
        for (int i = 0; i < m; i++) {
            int a = s.nextInt();
            int b = s.nextInt();
            int c = s.nextInt();
            if (sal.containsKey(b)) {
                sal.put(b, Math.min(sal.get(b), c));
            } else {
                sal.put(b, c);
            }
        }
        int result = 0;
        for (Map.Entry<Integer, Integer> en : sal.entrySet()) {
            result += en.getValue();
        }
        if (sal.size() < n - 1) {
            System.out.println(-1);
        } else {
            System.out.println(result);
        }

    }

}
