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.io.*;
import java.util.*;

public class Hierarchy
{
	public Scanner in = new Scanner(System.in);
	public PrintStream out = System.out;

	public int n, En;
	public int[] q;
	public long[][] G;
	
	public long oo = 987654321987654321L;
	
	public long[] cost;
	public boolean[] pick;
	
	public void main()
	{
		int i, j, k, x, y, c;
		
		n = in.nextInt();
		q = new int[n];
		
		TreeSet<Person> Q = new TreeSet<Person>();
		for(i=0;i<n;++i) 
		{
			q[i] = in.nextInt();
			Q.add(new Person(i, q[i]));
		}
		
		G = new long[n][n];
		init(G, oo);
		
		En = in.nextInt();
		for(k=0;k<En;++k)
		{
			x = in.nextInt()-1;
			y = in.nextInt()-1;
			c = in.nextInt();
			
			G[x][y] = Math.min(c, G[x][y]);
		}
				
		cost = new long[n];		
		Arrays.fill(cost, oo);
		
		Person who = Q.first();
		Q.remove(who);
		
		cost[who.i] = 0;		
		while(true)
		{
			//relax
			x = who.i;
			for(y=0;y<n;++y) if(G[x][y] < cost[y])
			{
				cost[y] = G[x][y];
			}
			
			//next
			if(Q.isEmpty()) break;
			who = Q.first();
			Q.remove(who);
		}
		
		long res = 0;
		for(x=0;x<n;++x) if(cost[x] >= oo)
		{
			res = -1; break;
		}
		else
		{
			res += cost[x];
		}
		
		out.println(res);
	}//end public void main()

	private class Person implements Comparable<Person>
	{
		public int i, qi;
		
		public Person(int ii, int qii) { i=ii; qi=qii;}
		
		public int compareTo(Person u)
		{
			if(qi != u.qi) return -(qi-u.qi);
			return i - u.i;
		}
	}
	
	public void init(long[][] A, long val)
	{ 
		for(int i=0;i<A.length;++i) Arrays.fill(A[i], val); 
	}	
	
	public static void main(String[] args)
	{
		(new Hierarchy()).main();
	}
}

