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 java.lang.Exception {
		BufferedReader kek = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter outkek = new PrintWriter(System.out);
		
		int count = 0;
		int sum = 0;

		int N = Integer.parseInt(kek.readLine());
		kek.readLine();
		//String[] input = kek.readLine().split(" "); Qualification is already known and Qai > Qbi is always true
		//int[] Qi = new int[N];						Therefore, we don't actually need to track it

		int[] costArray = new int[N + 1];
		
		for(int i = 1; i <= N; i++){
			costArray[i] = Integer.MAX_VALUE;	//CostArray will be used to track cost and if all workers are
		}										//included in hierarchy
		
		int M = Integer.parseInt(kek.readLine());

		
		for(int i = 0; i < M; i++){		//If we don't need to verify qualifications, we simply need to track
			String[] input = kek.readLine().split(" ");	// minimum cost of application and if valid hierarchy.
			// int A = Integer.parseInt(input[0]); Don't need this value.
			int B = Integer.parseInt(input[1]);
			int C = Integer.parseInt(input[2]);
			
			if(costArray[B] > C){
				costArray[B] = C; //This will not only track if index B has been used, but also the cost
			}
		}
		
		for(int i = 1; i <= N; i++){
			if(costArray[i] == Integer.MAX_VALUE){ //Has index been used? There should only be one unused
				count++;							// index, the top-supervisor. Any more and the hierarchy
			} else {								// is invalid.
				sum += costArray[i]; 		//If index used, sum up cost.
			}
		}
		
		if( count == 1 ){		//Again, if count is not equal to one we did not include every
			outkek.println(sum);// worker in hierarchy
		} else {
			outkek.println(-1);
		}
		outkek.close();
	}
	
	
	
}
