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;

public class B_Hierarchy {

    //15:58
    
    static int n, m;
    static int[] q, a, b, c;
    
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        
        n = s.nextInt();
        q = new int[n];

        for (int i = 0; i < n; i++) {
            q[i] = s.nextInt();
        }
        
        m = s.nextInt();
        a = new int[m];
        b = new int[m];
        c = new int[m];
        for (int i = 0; i < m; i++) {
            a[i] = s.nextInt() - 1;
            b[i] = s.nextInt() - 1;
            c[i] = s.nextInt();
        }
        
        int[] cs = new int[n];
        int rootcnt = 0;
        for (int i = 0; i < n; i++) {
            int mincost = Integer.MAX_VALUE;
            //search applications to i
            for (int j = 0; j < m; j++) {
                if(b[j] == i){
                    if(c[j] < mincost) mincost = c[j];
                }
            }
            if(mincost == Integer.MAX_VALUE){
                rootcnt++;
                if(rootcnt >= 2){
                    System.out.println(-1);
                    return;
                }
            }
            cs[i] = mincost;
        }
        

        int sum = 0;
        for (int i = 0; i < n; i++) {
            if(cs[i] == Integer.MAX_VALUE) continue;
            sum += cs[i];
        }
        System.out.println(sum);
    }

}
