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.util.*;
import java.lang.*;
import java.math.*;
import java.io.*;

public class b {
	
	int n, m;
	int[] q;
	boolean[] used;
	int[][] canW;
	
	private void solve() throws Exception {
		n = nextInt();
		q = new int[n];
		for (int i = 0; i < n; ++i)
			q[i] = nextInt();
		used = new boolean[n];
		canW = new int[n][n];
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < n; ++j)
				canW[i][j] = Integer.MAX_VALUE;
		m = nextInt();
		for (int i = 0; i < m; ++i) {
			int st = nextInt() - 1, en = nextInt() - 1, cost = nextInt();
			canW[st][en] = Math.min(canW[st][en], cost);
		}
		long res = 0;
		for (int i = 0; i < n; ++i) {
			int max = Integer.MIN_VALUE, ind = -1;
			for (int j = 0; j < n; ++j)
				if (!used[j] && q[j] > max) {
					max = q[j];
					ind = j;
				}
			if (i == 0)
				used[ind] = true;
			else {
				int min = Integer.MAX_VALUE;
				for (int j = 0; j < n; ++j)
					if (used[j] && canW[j][ind] < min) {
						min = canW[j][ind];
					}
				if (min == Integer.MAX_VALUE){
					out.print(-1);
					return;
				}
				else{
					res += min;
					used[ind] = true;
				}
			}
		}
		out.print(res);
	}

	public void run() {
		try {
			solve();
		} catch (Exception e) {
			NOO(e);
		} finally {
			out.close();
		}
	}

	PrintWriter out;
	BufferedReader in;
	StringTokenizer St;

	void NOO(Exception e) {
		e.printStackTrace();
		System.exit(1);
	}

	int nextInt() {
		return Integer.parseInt(nextToken());
	}

	long nextLong() {
		return Long.parseLong(nextToken());
	}

	double nextDouble() {
		return Double.parseDouble(nextToken());
	}

	String nextToken() {
		while (!St.hasMoreTokens()) {
			try {
				String line = in.readLine();
				St = new StringTokenizer(line);
			} catch (Exception e) {
				NOO(e);
			}
		}
		return St.nextToken();
	}

	private b(String name) {
		try {
			in = new BufferedReader(new FileReader(name + ".in"));
			St = new StringTokenizer("");
			out = new PrintWriter(new FileWriter(name + ".out"));
		} catch (Exception e) {
			NOO(e);
		}
	}

	private b() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			St = new StringTokenizer("");
			out = new PrintWriter(System.out);
		} catch (Exception e) {
			NOO(e);
		}
	}

	public static void main(String[] args) {
		Locale.setDefault(Locale.US);
		new b().run();
	}

}
