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 Main{
	public static void main(String args[]) throws Exception{
		BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
		int emp=Integer.parseInt(in.readLine());
		String q=in.readLine();
		
		Employee[] emplys=new Employee[emp];
		for(int i=0;i<emp;i++)
			emplys[i]=new Employee(i);
		
		int app=Integer.parseInt(in.readLine());
		for(int i=0;i<app;i++){
			String[] apps=in.readLine().split(" ");
			int f=Integer.parseInt(apps[0])-1;
			int t=Integer.parseInt(apps[1])-1;
			int w=Integer.parseInt(apps[2]);
			
			emplys[t].pq.add(new Edge(f,t,w));
		}
		
		int sum=0;
		int rootcount=0;
		for(int i=0;i<emp;i++){
			if(emplys[i].pq.isEmpty()){
				rootcount++;
				if(rootcount>1){
					sum=-1;
					break;
				}
			}
			else sum+=emplys[i].pq.poll().w;
		}
		
		System.out.println(sum);
	}
	
	static class Employee{
		PriorityQueue<Edge> pq;
		Edge real;
		int ind;
		
		public Employee(int ind){
			pq=new PriorityQueue<Edge>();
			this.ind=ind;
		}
	}
	
	static class Edge implements Comparable<Edge>{
		int w;
		int from, to;
		
		// public Edge(int w){
			// this.w=w;
		// }
		
		public Edge(int from, int to, int w){
			this.from=from;
			this.to=to;
			this.w=w;
		}
		
		public int compareTo(Edge e){
			return w-e.w;
		}
	}
}