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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;

public class P6 {

    static class Emp implements Comparable<Emp> {
        static int globId = 0;
        int id;
        int qual;
        int best = Integer.MAX_VALUE;

        public Emp(int qual) {
            this.qual = qual;
            id = globId++;
        }

        @Override
        public int compareTo(Emp o) {
            if (qual == o.qual)
                return id < o.id ? -1 : 1;
            return qual < o.qual ? -1 : 1;
        };
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int len = in.nextInt();
        ArrayList<Emp> emps = new ArrayList<Emp>();
        int[] vals = new int[len];
        Arrays.fill(vals, Integer.MAX_VALUE);
        for (int i = 0; i < len; ++i) {
            emps.add(new Emp(in.nextInt()));
        }
        int apps = in.nextInt();
        for (int i = 0; i < apps; ++i) {

            in.nextInt();
            int dest = in.nextInt() - 1;
            int cost = in.nextInt();
            Emp emp = emps.get(dest);
            emp.best = Math.min(emp.best, cost);
        }
        Collections.sort(emps);
        long cost = 0;
        for (int i = 0; i < emps.size() - 1; ++i) {
            if (emps.get(i).best < Integer.MAX_VALUE)
                cost += emps.get(i).best;
            else {
                cost = -1;
                break;
            }

        }
        out.println(cost);

        out.flush();
        out.close();
    }
}
