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 {
	
	final int infinity = 1000000010;

	public void _main() throws IOException {
		int n = nextInt();
		int[] q = new int[n];
		for (int i = 0; i < n; i++)
			q[i] = nextInt();
		
		int[] p = new int[n];
		for (int i = 0; i < n; i++)
			p[i] = i;
		for (int i = 0; i < n; i++)
			for (int j = i + 1; j < n; j++)
				if (q[p[i]] < q[p[j]]) {
					int t = p[i];
					p[i] = p[j];
					p[j] = t;
				}
		
		int[] rp = new int[n];
		for (int i = 0; i < n; i++)
			rp[p[i]] = i;
		p = rp;

		int m = nextInt();
		int[][] cost = new int[n][n];
		for (int[] arr : cost)
			Arrays.fill(arr, infinity);
		for (int i = 0; i < m; i++) {
			int a = nextInt() - 1;
			int b = nextInt() - 1;
			int c = nextInt();
			a = p[a];
			b = p[b];
			cost[a][b] = Math.min(cost[a][b], c);
		}
		int res = 0;
		for (int i = 1; i < n; i++) {
			int cur = infinity;
			for (int j = 0; j < i; j++)
				cur = Math.min(cur, cost[j][i]);
			if (cur >= infinity) {
				res = infinity;
				break;
			}
			res += cur;
		}
		out.print(res >= infinity ? -1 : res);
	}

	private BufferedReader in;
	private PrintWriter out;
	private StringTokenizer st;

	private String next() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			String rl = in.readLine();
			if (rl == null)
				return null;
			st = new StringTokenizer(rl);
		}
		return st.nextToken();
	}

	private int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	private long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	private double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	public static void main(String[] args) {
		new Thread(new Main()).start();
	}

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);

			_main();

			out.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(202);
		}
	}

}
