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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Hierarchy {
	public static class Employee{
		int id;
		int cost;
		public Employee(int id , int cost){
			this.id=id;
			this.cost=cost;
		}
	}
	public static void main(String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out,true);
		int N = Integer.parseInt(br.readLine());
		HashMap<Integer,Employee> graph = new HashMap<Integer,Employee>();
		for (int i = 0; i < N; i++) {
			graph.put(i, null);
		}
		br.readLine();
		int M = Integer.parseInt(br.readLine());
		for (int i = 0; i < M; i++) {
			String [] s = br.readLine().split("\\s+");
			int a = Integer.parseInt(s[0]);
			int b = Integer.parseInt(s[1]);
			int c = Integer.parseInt(s[2]);
			
			if(graph.get(b-1)==null)
				graph.put(b-1,new Employee(a-1,c));
			else{
				int cost = graph.get(b-1).cost;
				if(c<cost)
					graph.put(b-1,new Employee(a-1,c));
			}
		}
		int count = 0;
		int sum = 0;
		for (int i = 0; i < N; i++) {
			if(graph.get(i)==null)
				count++;
			else
				sum+=graph.get(i).cost;
		}
		
		pw.println(count>1?-1:sum);
		pw.flush();
	}
}
