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.*;

public class S17{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int[] ar=new int[n];
		for(int i=0;i<n;i++) ar[i]=sc.nextInt();

	    int m=sc.nextInt();
	    int[] par=new int[n+1];
	    for(int i=1;i<=n;i++){
	    	par[i]=i;
	    }
	    Pair[] pr=new Pair[m];
	    for(int i=0;i<m;i++){
	    	int a=sc.nextInt();
	    	int b=sc.nextInt(); 
	    	int c=sc.nextInt();
	    	pr[i]=new Pair(a,b,c);
          }
          Arrays.sort(pr,new SortPair());
          int ans=0;
          for(int i=0;i<m;i++){
              int a=pr[i].x;
              int b=pr[i].y;
              int c=pr[i].z;
              
              if(find(b,par)==b){
              	par[b]=a;
              	ans+=c;
              }
          }
          int cnt=0;
         for(int i=1;i<=n;i++){
            if(find(i,par)==i) cnt++;
         }
         if(cnt>1) System.out.println("-1");
         else System.out.println(ans);

	}
	static int find(int n,int[] par){
		if(n==par[n]){
			return n;
		}
		return par[n]=find(par[n],par);
	}
}

class Pair{
	int x,y,z;
	Pair(int x,int y, int z){
		this.x=x;
		this.y=y;
		this.z=z;
	}
}
class SortPair implements Comparator<Pair>{
	public int compare(Pair p1,Pair p2){
		return p1.z-p2.z;
	}
}