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.Collections;
import java.util.LinkedList;
import java.util.Scanner;

public class P17B
{
    private static int[] qual;

    public static void main(String[] args)
    {
        Scanner in = new Scanner(System.in);

        int n = in.nextInt();
        qual = new int[n];
        int maxQual = 0;
        for (int i = 0; i < n; i++)
        {
            qual[i] = in.nextInt();
            if (qual[i] > qual[maxQual])
                maxQual = i;
        }

        int m = in.nextInt();
        LinkedList<Edge> edges = new LinkedList<>();
        for (int i = 0; i < m; i++)
        {
            int a = in.nextInt() - 1;
            int b = in.nextInt() - 1;
            int c = in.nextInt();

            edges.add(new Edge(a, b, c));
        }
        Collections.sort(edges);

        int[] minVals = new int[n];
        for (int i = 0; i < n; i++)
            minVals[i] = i == maxQual ? 0 : -1;

        for (Edge e : edges)
        {
            if (minVals[e.d] == -1 || e.c < minVals[e.d])
                minVals[e.d] = e.c;
        }

        boolean exists = true;
        int sum = 0;
        for (int i = 0; i < n && exists; i++)
        {
            if (minVals[i] == -1)
            {
                sum = -1;
                exists = false;
            }
            else
                sum += minVals[i];
        }

        System.out.println(sum);
    }

    private static class Edge implements Comparable<Edge>
    {
        int s, d, c;

        public Edge(int s, int d, int c)
        {
            this.s = s;
            this.d = d;
            this.c = c;
        }

        @Override
        public int compareTo(Edge e)
        {
            if (qual[this.d] == qual[e.d])
                return this.c - e.c;
            return qual[e.d] - qual[this.d];

        }
    }
}
