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.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
/**
 * Erasyl Abenov
 * 
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
            try (PrintWriter out = new PrintWriter(outputStream)) {
                Task solver = new Task();
                solver.solve(1, in, out);
            }
    }
}
  
class Task {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n = in.nextInt();
        boolean over[] = new boolean[n];
        int price[] = new int[n];
        for(int i = 0; i < n; ++i){
            int qualification = in.nextInt();
        }
        int m = in.nextInt();
        for(int i = 0; i < m; ++i){
            int a = in.nextInt();
            int b = in.nextInt() - 1;
            int c = in.nextInt();
            if(over[b]) price[b] = Math.min(price[b], c);
            else{
                over[b] = true;
                price[b] = c;
            }
        }
        int answer = 0;
        int  nonQual = 0;
        for(int i = 0; i < n; ++i){
            if(!over[i]) nonQual++;
            else answer += price[i];
        }
        if(nonQual > 1) answer = -1;
        out.println(answer);
    }
}

class InputReader {
    private final BufferedReader reader;
    private StringTokenizer tokenizer;
  
    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
        tokenizer = null;
    }
    
    public String nextLine() {
        try {
            return reader.readLine();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }
  
    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(nextLine());
        }
        return tokenizer.nextToken();
    }
    public double nextDouble() {
        return Double.valueOf(next());
    }
    
    
    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong() {
        return Long.parseLong(next());
    }
    public BigInteger nextBig() {
        return new BigInteger(next());
    }
}
 class IOUtils {
    public static void nextIntArrays(InputReader in, int[]... arrays) {
        for (int i = 0; i < arrays[0].length; i++) {
            for (int[] array : arrays) {
                array[i] = in.nextInt();
            }
        }
    }
}