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.
//package main;

import java.io.*;
import java.util.*;

public class Main{
    public static void main(String[] args) throws IOException {
        Reader.init(System.in);
        
        int N = Reader.nextInt();
        int[] q = new int[N];
        //DisjointSet set = new DisjointSet(N);
        int[][] cost = new int[N][N];
        TreeSet<Employee> emp = new TreeSet<Employee>();
        int total = 0;
        
        for(int i = 0; i < N; i++) {
            q[i] = Reader.nextInt();
            emp.add(new Employee(q[i], i));
        }
        
        int A = Reader.nextInt();
        for(int i = 0; i < A; i++){
            int a = Reader.nextInt()-1;
            int b = Reader.nextInt()-1;
            if(cost[a][b] == 0) {
                cost[a][b] = Reader.nextInt()+1;
            } else {
                cost[a][b] = Math.min(Reader.nextInt()+1, cost[a][b]);
            }
        }
        if(A < N-1) {
           System.out.print(-1);
           return;
        }
        
        boolean fail = false;
        for(Employee e : emp) {
            if(fail) {
                System.out.print(-1);
                return;
            }
            int iMin = -1;
            for(int i = 0; i < N; i++) {
                if(cost[i][e.i] > 0) {
                    if(iMin == -1) {
                        iMin = i;
                    } else if(cost[i][e.i] < cost[iMin][e.i]) {
                        iMin = i;
                    }
                }
            }
            if(iMin == -1) {
                fail = true;
            }
            else{
                total+=(cost[iMin][e.i] -1);
            }
        }
        
        System.out.print(total);
    }
}
class Employee implements Comparable<Employee>{
    int q, i;
    Employee(int qualification, int index){
        q = qualification;
        i = index;
    }
    @Override public int compareTo(Employee that) {
        final int BEFORE = -1;
        final int EQUAL = 0;
        final int AFTER = 1;

        //this optimization is usually worthwhile, and can
        //always be added
        if (this == that) return EQUAL;
        
        if(this.q < that.q) return BEFORE;
        if(this.q > that.q) return AFTER;
        
        if(this.i < that.i) return BEFORE;
        if(this.i > that.i) return AFTER;
        
        return EQUAL;
    }
}

// Class for buffered reading int and double values 
// From: http://www.cpe.ku.ac.th/~jim/java-io.html
class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;

    //call this method to initialize reader for InputStream
    static void init(InputStream input) {
        reader = new BufferedReader(
                     new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }

    //get next word
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                   reader.readLine() );
        }
        return tokenizer.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
	
    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
}