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.Scanner;
import java.util.ArrayList;

public class main{
    public static void main(String []args){
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        ArrayList<Integer> list = new ArrayList<Integer>();
        int[] qual = new int[n];
        for(int i=0;i<n;i++){
            list.add(i+1);
            qual[i]=s.nextInt();
        }
        int m = s.nextInt();
        int[][] appl = new int[m][3];
        for(int i = 0;i<m;i++){
            appl[i][0] = s.nextInt();
            appl[i][1] = s.nextInt();
            appl[i][2] = s.nextInt();
        }
        for(int i = 0;i<m;i++){
            for(int j = i+1;j<m;j++){
                if(appl[i][2]>appl[j][2]){
                    int[] temp = appl[i];
                    appl[i] = appl[j];
                    appl[j] = temp;
                }
            }
        }
        int cost = 0;
        for(int i = 0;i<m;i++){
            if(list.size()==1){
                break;
            }
            if(list.indexOf(appl[i][1])!=-1){
                cost+= appl[i][2];
                list.remove(list.indexOf(appl[i][1]));
            }
        }
        if(list.size()==1) System.out.println(cost);
        else System.out.println("-1");
    }
    
}
   