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.*;

/**
 * Created by HREN_VAM.
 */

public class B implements Runnable{

    BufferedReader in;
    PrintWriter out;
    StringTokenizer st;
    public static final String filename = "b";

    public static int MAX = 1000000000;

    class I implements Comparable<I>{

        int val;
        int num;


        I(int val, int num){
            this.val = val;
            this.num = num;
        }
        
        public int compareTo(I o){
            return Double.compare(val, o.val);
        }
    }
    
    public void solve() throws IOException{
        int n = nextInt();
        int[][] a = new int[n][n];
        I[] b = new I[n];

        for(int i = 0;i < n;i ++){
            b[i] = new I(nextInt(), i);
        }
        Arrays.sort(b);
        
        for(int i = 0;i < n;i ++){
            for(int j = 0;j < n;j ++){
                a[i][j] = MAX;
            }
        }

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

        int res = 0;
        for(int i = 0;i < n - 1;i ++){
            int min = MAX;
            for(int j = 0;j < n;j ++){
                min = Math.min(min, a[b[i].num][j]);
            }
            if(min == MAX){
                out.println(-1);
                return;
            }
            res += min;
        }
        out.println(res);
    }

    public void run(){
        try{
            Locale.setDefault(Locale.US);
            in = new BufferedReader(new InputStreamReader(System.in));
            //in = new BufferedReader(new FileReader(filename + ".in"));
            out = new PrintWriter(System.out);
            //out = new PrintWriter(new FileWriter(filename + ".out"));
            st = new StringTokenizer("");
            solve();
            out.close();
        } catch(IOException e){
            throw new RuntimeException(e);
        }
    }

    public static void main(String[] args){
        new Thread(new B()).start();
    }

    public String nextToken() throws IOException{
        while(!st.hasMoreTokens()){
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }

    public int nextInt() throws IOException{
        return Integer.parseInt(nextToken());
    }

    public double nextDouble() throws IOException{
        return Double.parseDouble(nextToken());
    }
}