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 Exception
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int v = Integer.parseInt(br.readLine()); 
		ArrayList<Node> graph = new ArrayList<Node>(); 
		for(int i=0; i<v; i++) graph.add(new Node(i+1)); 
		br.readLine(); 
		int e = Integer.parseInt(br.readLine()); 
		for(int j=0; j<e; j++)
		{
			String input = br.readLine(); 
			StringTokenizer tk = new StringTokenizer(input); 
			Node s = graph.get(Integer.parseInt(tk.nextToken())-1);
			Node d = graph.get(Integer.parseInt(tk.nextToken())-1);
			int weight = Integer.parseInt(tk.nextToken()); 
			if(weight<d.min)
			{
				d.min=weight; 
			}
		}
		int num = 0; 
		int sum = 0; 
		for(int i=0; i<graph.size(); i++)
		{
			Node temp = graph.get(i);
			if(temp.min!=Integer.MAX_VALUE)
				sum+=temp.min; 
			else num++;
		}
		if(num==1) System.out.println(sum);
		else System.out.println("-1");
	}
}

class Node
{
	int name; 
	ArrayList<Edge> conns; 
	int min;
	public Node(int name)
	{
		this.name=name;
		conns=new ArrayList<Edge>(); 
		min=Integer.MAX_VALUE;
	}
}

class Edge implements Comparable<Edge>
{
	Node s,d;
	int weight; 
	public Edge(Node s, Node d, int weight)
	{
		this.s=s;
		this.d=d;
		this.weight=weight; 
	}
	
	public int compareTo(Edge other)
	{
		return weight - other.weight; 
	}
}