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 Hierarchy {

	static int[] a = null;

	static void solve() {
		Scanner sc = new Scanner(System.in);
		a = new int[10001];
		int n = sc.nextInt(), x=0, y=0, m=0;
		for (int i=0; i<=n; i++) {
			m=sc.nextInt();
			a[i]=1000000000;
		}
		for (int i=1; i<=m; i++) {
			sc.nextInt(); 
			x = sc.nextInt(); y = sc.nextInt();
			a[x] = Math.min(a[x],y);
		}
		x=y=0;
		for (int i=1; i<=n; i++) {
			if (a[i]!=1000000000) {
				x++; y+=a[i];
			}
		}
		if (n<x+2) System.out.println(y);
		else System.out.println(-1);
		sc.close();
	}

	public static void main(String[] args) {
		solve();
	}
}