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.Arrays;
import java.util.Scanner;

public class b17 {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		for (int i = 0; i < n; i++) {
			in.next();
		}
		long[] best = new long[n];
		Arrays.fill(best, Long.MAX_VALUE);
		int m = in.nextInt();
		for (int i = 0; i < m; i++) {
			in.next();
			int idx = in.nextInt()-1;
			long cost = in.nextLong();
			best[idx] = Math.min(best[idx], cost);
		}
		long total = 0;
		int numUnpossible = 0;
		long worstPossible = 0;

		for (int i = 0; i < best.length; i++) {
			if (best[i] == Long.MAX_VALUE) {
				numUnpossible++;
			} else {
				total += best[i];
				worstPossible = Math.max(worstPossible, best[i]);
			}
		}
		if (numUnpossible == 0)
			total -= worstPossible;
		if (numUnpossible > 1)
			System.out.println(-1);
		else
			System.out.println(total);

	}
}