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.lang.*;
import java.math.*;
import java.io.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;

public class _P017B{
    Scanner sc=new Scanner(System.in);

    int INF=1<<28;
    double EPS=1e-9;

    int n;
    int[] q;
    int m;
    int[] a, b, c;

    void run(){
        n=sc.nextInt();
        q=new int[n];
        for(int i=0; i<n; i++){
            q[i]=sc.nextInt();
        }
        m=sc.nextInt();
        a=new int[m];
        b=new int[m];
        c=new int[m];
        for(int i=0; i<m; i++){
            a[i]=sc.nextInt()-1;
            b[i]=sc.nextInt()-1;
            c[i]=sc.nextInt();
        }
        solve();
    }

    void solve(){
        int[] costs=new int[n];
        fill(costs, INF);
        for(int i=0; i<m; i++){
            costs[b[i]]=min(costs[b[i]], c[i]);
        }
        int sum=0;
        int count=0;
        for(int i=0; i<n; i++){
            if(costs[i]<INF){
                sum+=costs[i];
            }else{
                count++;
            }
        }
        println((count==1?sum:-1)+"");
    }

    void println(String s){
        System.out.println(s);
    }

    void print(String s){
        System.out.print(s);
    }

    void debug(Object... os){
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args){
        Locale.setDefault(Locale.US);
        new _P017B().run();
    }
}
