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 test 
{  
    public static void main(String[] args) 
    {
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();

        int root=0;
        int tmp=0;
        for(int i=0;i<n;i++)
        {
            int a=in.nextInt();
            if(a>tmp)
            {
        	root=i;
        	tmp=a;
            }
        }
        
        int[][] matrix=new int[n][n];
        for(int i=0;i<n;i++)
            Arrays.fill(matrix[i],Integer.MAX_VALUE);
        
        int m=in.nextInt();
        for(int i=0;i<m;i++)
        {
            int a=in.nextInt()-1;
            int b=in.nextInt()-1;
            int c=in.nextInt();
            
            matrix[a][b]=Math.min(matrix[a][b],c);
        }
        
        int total=0;
        
        for(int i=0;i<n;i++)
        {
            if(i==root)
        	continue;
            
            int min=Integer.MAX_VALUE;
            
            for(int j=0;j<n;j++)
            {
        	min=Math.min(min,matrix[j][i]);
            }
            
            if(min==Integer.MAX_VALUE)
            {
        	System.out.println(-1);
        	return;
            }
            else
        	total+=min;
        }

        System.out.println(total);
        
    }
}




