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.Scanner;
public class Tester {

      public static void main(String args[])
	  {     Scanner sc=new Scanner(System.in);
			int noe=sc.nextInt();
			sc.nextLine();
			String Qual[]=sc.nextLine().split(" ");
			int noa=sc.nextInt();
			sc.nextLine();
			int a[]=new int[noa];
			int b[]=new int[noa];
			int c[]=new int[noa];
			int q[]=new int[noa];
			int maxq=0,maxqi=0;
			for(int i=0;i<noe;i++)
			{
				if(Integer.parseInt(Qual[i])>maxq)
				{	maxq=Integer.parseInt(Qual[i]);
					maxqi=i;}
			}
			//System.out.println(maxqi);
			for(int i=0;i<noa;i++)
			{
				String temp[]=sc.nextLine().split(" ");
				a[i]=Integer.parseInt(temp[0]);
				b[i]=Integer.parseInt(temp[1]);
				c[i]=Integer.parseInt(temp[2]);
			
			}
			int count=0,cost=0;
			for(int i=0;i<noe;i++)
			{   if(i!=maxqi)
				{int coh=0,cp=0;
				for(int j=0;j<noa;j++)
				{
					if(b[j]==i+1 && cp==0)
						{coh=c[j];
						cp++;}
					else if(b[j]==i+1 && coh>c[j])
						coh=c[j];
					
				}
				cost=cost+coh;
				if(cp!=0)
					count++;
				}
			}
			if(count==noe-1)
				System.out.println(cost);
			else
				System.out.println("-1");
	  }
		
}
