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.Arrays;
import java.util.Scanner;


public class p17b {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        
        emp[] e = new emp[n];
        for (int i = 0; i < e.length; i++) {
            e[i] = new emp(in.nextInt(),i);
        }
        Arrays.sort(e);
        int m = in.nextInt();
        
        int[][] all = new int[n][n];
        for (int i = 0; i < all.length; i++) {
            Arrays.fill(all[i], 9999999);
        }
        for (int i = 0; i < m; i++) {
            int a = in.nextInt()-1;
            int b = in.nextInt()-1;
            int c = in.nextInt();
            
            all[b][a] = Math.min(all[b][a], c);
        }
        
        int cost  = 0;
        for (int i = 1; i < e.length; i++) {
            int index = e[i].index;
            int min = 9999999;
            for (int j = 0; j < all[index].length; j++) {
                min = Math.min(min, all[index][j]);
            }
            if(min == 9999999) {
                System.out.println(-1);
                return;
            }
            cost += min; 
        }
        System.out.println(cost);
        
    }
}
class emp implements Comparable<emp>{
    int q;
    int index;
    public emp(int a , int b) {
        q = a;
        index = b;
    }
    public int compareTo(emp o) {
        
        return o.q-q;
    }
}