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.*;
public class Main
{
	public static void main(String args[]) throws IOException
	{
		Scanner sc = new Scanner(System.in);		
		int n = sc.nextInt();
		boolean boss[] = new boolean[n];
		int min[] = new int[n];
		for(int i=0;i<n;i++)
		{
			sc.nextInt();
			boss[i] = false;
		}
		int m = sc.nextInt(), a, b, c;
		for(int i=0;i<m;i++)
		{
			a = sc.nextInt();
			b = sc.nextInt();
			c = sc.nextInt();
			if (!boss[b-1]) 
			{
				min[b-1] = c;
				boss[b-1] = true;
			}
			else min[b-1] = Math.min(min[b-1],c);			
		}
		int wboss = 0, sum = 0;
		for(int i=0;i<n;i++)
		{
			if (!boss[i]) wboss++;
			else sum += min[i];
		}
		if (wboss > 1) System.out.print("-1");  else System.out.print(sum);
	}
}