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 implements Runnable {
	
	public void run() {
		try {
			input = new BufferedReader(new InputStreamReader(System.in));
			output = new PrintWriter(System.out);
			tokenizer = new StreamTokenizer(input);
			
				int n = nextInt();
				int m = n;
				int[] map = new int[n];
				int count = n - 1;
				Arrays.fill(map, -1);

				while(m-- > 0)
					nextInt();

				m = nextInt();
				while(m-- > 0) {
					int a = nextInt() - 1,
						b = nextInt() - 1,
						c = nextInt();

					if(map[b] == -1) {
						map[b] = c;
						count--;
					} else {
						map[b] = Math.min(map[b], c);
					}
				}

				if(count > 0)
					output.println(-1);
				else {
					int ans = 0;

					for(int i = 0; i < n; i++)
						if(map[i] >= 0)
							ans += map[i];

					output.println(ans);
				}
				

			input.close();
			output.close();
		} catch(Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}
	
	BufferedReader input;
	PrintWriter output;
	StreamTokenizer tokenizer;
	
	int nextInt() throws IOException {
		tokenizer.nextToken();
		return (int)tokenizer.nval;
	}
	
	double nextDouble() throws IOException {
		tokenizer.nextToken();
		return tokenizer.nval;
	}
	
	String nextWord() throws IOException {
		tokenizer.nextToken();
		return tokenizer.sval;
	}
	
	String nextLine() throws IOException {
		return input.readLine();
	}
	
	public static void main(String[] args) {
		new Main().run();
	}
}
