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.*;
import java.io.*;

public class B17{
	public static void main(String ar[])throws IOException{
		Scanner sc=new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter pw=new PrintWriter(System.out);
		int n=sc.nextInt();
		long emp[]=new long[n+1];
		for(int i=1;i<=n;i++)
			emp[i]=sc.nextLong();
		int m=sc.nextInt();
		int arr[][]=new int[n+1][n+1];
		for(int i=0;i<=n;i++){
			for(int j=0;j<=n;j++){
				arr[i][j]=1000001;
			}
		}
		try{
			for(int i=0;i<m;i++){
				int a=sc.nextInt();
				int b=sc.nextInt();
				int c=sc.nextInt();
				if(emp[a]>emp[b]){
					arr[a][b]=Math.min(c,arr[a][b]);
				}
			}
			int head=0;
			int no=0;
			for(int i=1;i<=n;i++){
				for(int j=1;j<=n;j++){
					if(arr[j][i]!=1000001)
						break;
					if(j==n){
						head=i;
						no++;
					}
				}
			}
			// for(int i=1;i<=n;i++){
			// 	for(int j=1;j<=n;j++){
			// 		System.out.print(arr[i][j]);
			// 	}
			// 	System.out.println();
			// }
			if(no>1){
				System.out.println(-1);
				throw new Exception();
			}
			long cont=0;
			for(int i=1;i<=n;i++){
				if(i!=head){
					int min=1000001;
					for(int j=1;j<=n;j++)
						min=Math.min(min,arr[j][i]);
					cont+=min;
				}
			}
			pw.println(cont);
		}
		catch(Exception e){
			int zz=0;
		}
		finally{
			pw.close();
		}
	}
}