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.io.*;
import java.util.*;
/**
 * ecnuacm
 * Main.java
 */
public class Main {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		
		int n,m;
		int[] a = new int[1001];
		n = in.nextInt();
		
		for(int i=1;i<=n;i++){
			int t = in.nextInt();
			a[i] = -1;
		}
		m = in.nextInt();
		for(int i=0;i<m;i++){
			int aa,b,c;
			aa = in.nextInt();
			b = in.nextInt();
			c = in.nextInt();
			if(a[b] == -1)a[b] = c;
			else a[b]=Math.min(c, a[b]);
		}
		int cnt = 0,ans=0;
		for(int i=1;i<=n;i++){
			if(a[i] == -1)cnt++;
			else ans += a[i];
		}
		if(cnt != 1)ans = -1;
		System.out.printf("%d\n", ans);
	}
}

 	 				  		 		    	 	 	