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.
//package codeforce;

import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Problem17B {

	public static int[] costs;
	public static int total_cost =0;
	
	public static void main(String[] args) throws java.lang.Exception {
		BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(buffer.readLine().trim()); // Number of employees
		String[] input = buffer.readLine().trim().split("\\s+");
				
		costs = new int[n+1];
		
		for(int i=0; i<=n; i++){
			costs[i] = Integer.MAX_VALUE;
		}
		
		int m = Integer.parseInt(buffer.readLine().trim());
		
		for(int i=0; i<m; i++){
			input = buffer.readLine().trim().split("\\s+");
			int a = Integer.parseInt(input[0]),
				b = Integer.parseInt(input[1]),
				c = Integer.parseInt(input[2]);
			
			costs[b] = Math.min(costs[b], c);
		}
		
		int gap=0;
		for(int i=1; i<=n; i++){
			if(costs[i] == Integer.MAX_VALUE)
				gap++;
			else
				total_cost += costs[i];
		}
		
		if(gap>1)
			System.out.println("-1");
		else
			System.out.println(total_cost);
		
		//Do DFS on qq using the cost table
//		dfs(max_q_idx);
	}	

}
