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 {
	static Scanner in;
	static PrintWriter out;

	public static void main(String[] args) throws Exception {
		in = new Scanner(System.in);
		out = new PrintWriter(System.out);
		
		int n = in.nextInt();
		int[] q = new int[n];
		for (int i = 0; i < n ;i++)
			q[i] = in.nextInt();

	    int m = in.nextInt();
		long[][] cost = new long[n][n];
		int inf = 1000000000;
		for (int i = 0; i < n; i++) Arrays.fill(cost[i], inf);
		for (int i = 0; i < m; i++) {
			int a = in.nextInt() - 1;
			int b = in.nextInt() - 1;
			int c = in.nextInt();
			cost[a][b] = Math.min(c, cost[a][b]);
		}

		int[] num = new int[n];
		for (int i = 0; i < n; i++)
			num[i] = ((1000000 - q[i]) << 10) + i;

		Arrays.sort(num);
		for (int i = 0; i < n; i++) num[i] &= (1 << 10) - 1;

		long[] min = new long[n];
		min[0] = 0;
		for (int i = 1; i < n; i++) {
			int num_j = 0;
			for (int j = 0; j < n; j++)
				if (cost[j][num[i]] < cost[num_j][num[i]]) num_j = j;

			min[i] = min[i - 1] + cost[num_j][num[i]];
		}

		if (min[n - 1] >= inf) out.println(-1);
		else out.println(min[n - 1]);

		out.close();
	}
}