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.*;

public class Beta17PB {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		new Beta17PB().solve();
	}
	
	int oo = 987654321;

	public void solve() {
		Scanner scan = new Scanner(System.in);
		int n;
		n = scan.nextInt();
		int[] qualification = new int[n];
		int[][] supervisor = new int[n][n];
		for(int i=0; i<n; i++) {
			qualification[i] = scan.nextInt();
			Arrays.fill(supervisor[i], -1);
		}
		int m, a, b, c;
		m = scan.nextInt();
		for(int i=0; i<m; i++) {
			a = scan.nextInt()-1; 
			b = scan.nextInt()-1;
			c = scan.nextInt();
			if(supervisor[a][b]<0) supervisor[a][b] = c;
			else supervisor[a][b] = Math.min(supervisor[a][b], c);
		}
		int cost = 0, base = 0, baseCost = -1;
		for(int i=0; i<n; i++) {
			int temp = oo;
			for(int j=0; j<n; j++) {
				if(j==i) continue;
				if(supervisor[j][i]>=0) temp = Math.min(temp, supervisor[j][i]);
			}
			if(temp<oo) {
				cost += temp;
				baseCost = Math.max(baseCost, temp);
			} else {
				base ++;
			}
		}
		if(base>1) {
			System.out.println(-1);
		} else if(base==1) {
			System.out.println(cost);
		} else {
			System.out.println(cost-baseCost);
		}
	}
}
