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 {
	private BufferedReader in;
	private PrintWriter out;
	private StringTokenizer st;
	
	private void eat(String line)
	{
		st = new StringTokenizer(line);
	}
	
	private String next() throws IOException
	{
		while(!st.hasMoreTokens()) {
			String line = in.readLine();
			if(line == null)
				return null;
			eat(line);
		}
		return st.nextToken();
	}
	
	private int nextInt() throws IOException
	{
		return Integer.parseInt(next());
	}
	
	public void run()
	{
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(new OutputStreamWriter(System.out));
			eat("");
			
			go();
			
			out.close();
		} catch(Exception e) {
			e.printStackTrace();
			System.exit(-1);
		}
	}
	
	public static void main(String[] args)
	{
		new Thread(new Main()).start();
	}
	
	public void go() throws IOException
	{
		int n = nextInt();
		int[] q = new int[n];
		for(int i = 0; i < n; ++i)
			q[i] = nextInt();
		int m = nextInt();
		int[] a = new int[m], b = new int[m], c = new int[m];
		for(int i = 0; i < m; ++i) {
			a[i] = nextInt() - 1;
			b[i] = nextInt() - 1;
			c[i] = nextInt();
		}
		int result = new Algo().solve(n, q, m, a, b, c);
		out.println(result);
	}
}

final class Algo
{
	private int[] getOrder(int n, final int[] q)
	{
		Integer[] order = new Integer[n];
		for(int i = 0; i < n; ++i)
			order[i] = i;
		Arrays.sort(order, new Comparator<Integer>() {
			public int compare(Integer a, Integer b)
			{
				return q[b] - q[a];
			}
		});
		int[] result = new int[n];
		for(int i = 0; i < n; ++i)
			result[i] = order[i];
		return result;
	}
	
	public int solve(int n, int[] q, int m, int[] a, int[] b, int[] c)
	{
		int[] order = getOrder(n, q);
		int[] weight = new int[n];
		Arrays.fill(weight, Integer.MAX_VALUE);
		weight[order[0]] = 0;
		
		int result = 0;
		boolean[] processed = new boolean[n];
		for(int i = 0; i < order.length; ++i) {
			int cur = order[i];
			processed[cur] = true;
			if(weight[cur] == Integer.MAX_VALUE)
				return -1;
			result += weight[cur];
			for(int j = 0; j < m; ++j)
				if(a[j] == cur && !processed[b[j]])
					weight[b[j]] = Math.min(weight[b[j]], c[j]);
		}
		return result;
	}
}
