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 Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        Node data[] = new Node[n];
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            int q = sc.nextInt();
            data[i] = new Node(i + 1, q);
            max = Math.max(max, q);
        }
        //Arrays.parallelSort(data);
        int m = sc.nextInt();
        if (m == 0 && n != 1) {
            System.out.println(-1);
            return;
        }
        int ap[][] = new int[m][3];
        for (int i = 0; i < m; i++) {
            ap[i][0] = sc.nextInt();
            ap[i][1] = sc.nextInt();
            ap[i][2] = sc.nextInt();
        }
        int ans = 0;
        int f = 1;
        for(int i = 0; i < n; i++) {
            if (data[i].q == max) {
                continue;
            }
            int w = Integer.MAX_VALUE;
            int flag = 0;
            for(int j=0; j < m; j++) {
                if(data[i].num == ap[j][1] && ap[j][2] < w) {
                    w = ap[j][2];
                    flag = 1;
                }
            }
            if(flag==0) {
                f = 0;
                break;
            }
            ans += w;
        }
        if (f == 0) {
            System.out.println(-1);
        } else {
            System.out.println(ans);
        }
    }

    public static class Node implements Comparable<Node> {
        int num;
        int q;

        public Node(int num, int q) {
            this.num = num;
            this.q = q;
        }

        @Override
        public int compareTo(Node o) {
            return this.q - o.q;
        }
    }
}
