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.*;
import java.io.*;

public class B0017 {

    public static void main(String args[]) throws Exception {
        new B0017();
    }

    public final long INF = 1000000000000L;

    B0017() throws Exception {
        PandaScanner sc = null;
        PrintWriter out = null;
        try {
            sc = new PandaScanner(System.in);
            out = new PrintWriter(System.out);
        } catch (Exception ignored) {
        }

        int n = sc.nextInt();
        for (int i = 0; i < n; i++) {
            sc.nextInt();
        }

        long[] cost = new long[n];
        Arrays.fill(cost, INF);

        int m = sc.nextInt();
        for (int i = 0; i < m; i++) {
            int a = sc.nextInt() - 1;
            int b = sc.nextInt() - 1;
            int c = sc.nextInt();
            cost[b] = Math.min(cost[b], c);
        }

        long res = 0;
        long max = 0;
        for (int i = 0; i < n; i++) {
            max = Math.max(max, cost[i]);
            res += cost[i];
        }
        res -= max;
        if (res >= INF) {
            out.println(-1);
        }
        else {
            out.println(res);
        }

        out.close();
        System.exit(0);
    }


    //The PandaScanner class, for Panda fast scanning!
    public class PandaScanner {
        BufferedReader br;
        StringTokenizer st;
        InputStream in;

        PandaScanner(InputStream in) throws Exception {
            br = new BufferedReader(new InputStreamReader(this.in = in));
        }

        public String next() throws Exception {
            if (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine().trim());
                return next();
            }
            return st.nextToken();
        }

        public boolean hasNext() throws Exception {
            return (st != null && st.hasMoreTokens()) || in.available() > 0;
        }

        public long nextLong() throws Exception {
            return Long.parseLong(next());
        }

        public int nextInt() throws Exception {
            return Integer.parseInt(next());
        }
    }
}
