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.awt.geom.Point2D;
import java.text.*;
import java.math.*;
import java.util.*;

public class Main implements Runnable {

	final String filename = "";

	public void solve() throws Exception {
		int n = iread();
		int[] q = new int[n];
		for (int i=0;i<n; i++)
			q[i] = iread();
		ArrayList<Integer>[] arr = new ArrayList[n];
		for (int i=0; i<n; i++)
			arr[i] = new ArrayList<Integer>();
		for (int m = iread(); m>0; m--)
		{
			int a = iread(), b = iread(), c = iread();
			a--;
			b--;
			arr[b].add(c);
		}
		int c = 0;
		for (int i=0;i<n;i++)
			if (arr[i].size()==0)
				c++;
		if (c>1)
		{
			out.write("-1");
			return;
		}
		long ans = 0;
		for (int i=0; i<n; i++)
		{
			if (arr[i].size()==0) continue;
			int min = arr[i].get(0);
			for (int j=1; j<arr[i].size(); j++)
				min = Math.min(min, arr[i].get(j));
			ans+=min;
		}
		out.write(ans+"\n");
	}

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new BufferedWriter(new OutputStreamWriter(System.out));
			// in = new BufferedReader(new FileReader(filename+".in"));
			// out = new BufferedWriter(new FileWriter(filename+".out"));
			solve();
			out.flush();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	public int iread() throws Exception {
		return Integer.parseInt(readword());
	}

	public double dread() throws Exception {
		return Double.parseDouble(readword());
	}

	public long lread() throws Exception {
		return Long.parseLong(readword());
	}

	BufferedReader in;

	BufferedWriter out;

	public String readword() throws IOException {
		StringBuilder b = new StringBuilder();
		int c;
		c = in.read();
		while (c >= 0 && c <= ' ')
			c = in.read();
		if (c < 0)
			return "";
		while (c > ' ') {
			b.append((char) c);
			c = in.read();
		}
		return b.toString();
	}

	public static void main(String[] args) {
		try {
			Locale.setDefault(Locale.US);
		} catch (Exception e) {

		}
		new Thread(new Main()).start();
		// new Thread(null, new Main(), "1", 1<<25).start();
	}
}