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.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class Hierarchy {
	static Node[] edgeList;
	int[] cost;
	Scanner input;

	public static void main(String[] args) {
		Hierarchy obj = new Hierarchy();
		obj.input = new Scanner(System.in);
		int n = obj.input.nextInt();
		int[] qualifications = new int[n];
		obj.cost = new int[n + 1];
		int result = 0;
		int numOfSup = 0;
		for (int i = 0; i < n; i++) {
			qualifications[i] = obj.input.nextInt();
		}
		int m = obj.input.nextInt();
		edgeList = new Node[m];
		for (int i = 1; i < n + 1; i++) {
			obj.cost[i] = -1;
		}
		obj.createEdgeList();
		obj.solve();
		for (int i = 1; i < n + 1; i++) {
			if (obj.cost[i] == -1) {
				numOfSup++;
			} else {
				result += obj.cost[i];
			}
		}
		if (numOfSup == 1) {
			System.out.println(result);
		} else {
			System.out.println(-1);
		}
	}

	private void createEdgeList() {
		int start, end, value;
		for (int i = 0; i < edgeList.length; i++) {
			start = input.nextInt();
			end = input.nextInt();
			value = input.nextInt();
			edgeList[i] = new Node(start, end, value);
		}
	}

	private void solve() {
		sort();
		int child;
		for (int i = 0; i < edgeList.length; i++) {
			child = edgeList[i].start;
			if (cost[child] == -1) {
				cost[child] = edgeList[i].value;
			}
		}

	}

	static void sort() {
		Arrays.sort(edgeList, new Comparator<Node>() {
			public int compare(Node n1, Node n2) {
				return n1.value - n2.value;
			}
		});
	}

	public class Node {
		int start, end, value;

		public Node(int end, int start, int value) {
			this.start = start;
			this.value = value;
			this.end = end;
		}
	}

}
