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.Locale;
import java.util.Scanner;

public class B{

    int n, m;
    int[] q, a, b, c;

    void exe(){
        Scanner sc=new Scanner(System.in);
        n=sc.nextInt();
        q=new int[n];
        for(int i=0; i<n; i++)
            q[i]=sc.nextInt();
        m=sc.nextInt();
        a=new int[m];
        b=new int[m];
        c=new int[m];
        for(int i=0; i<m; i++){
            a[i]=sc.nextInt();
            b[i]=sc.nextInt();
            c[i]=sc.nextInt();
        }
        int max=0;
        for(int i=0; i<n; i++){
            if(q[i]>q[max])
                max=i;
        }
        int cost=0;
        for(int i=0; i<n; i++){
            if(i==max)
                continue;
            int minj=-1;
            for(int j=0; j<m; j++){
                if(b[j]-1==i){
                    if(minj==-1||c[j]<c[minj]){
                        minj=j;
                    }
                }
            }
            if(minj==-1){
                System.out.println("-1");
                return;
            }
            cost+=c[minj];
        }
        System.out.println(""+cost);
    }

    public static void main(String[] args){
        Locale.setDefault(Locale.US);
        new B().exe();
    }
}
