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 {

	public int[] Parent;
	public int[] counter;
	public int total = 0;
	boolean[] visited;
	public int cnt;

	class edge {
		int supervisor;
		int employee;
		int cost;

	}

	void merge(int a, int b) {
		int ParentA = find(a);
		int ParentB = find(b);

		if (ParentA == ParentB) {
			return;

		}

		if (counter[a] >= counter[b]) {
			Parent[ParentB] = ParentA;
			counter[a] += counter[b];
		} else {
			Parent[ParentA] = ParentB;
			counter[b] += counter[a];
		}

		cnt--;
		
	}

	int find(int a) {
		if (a == Parent[a])
			return a;

		return Parent[a] = find(Parent[a]);
	}

	class Mycomp implements Comparator<Object> {
		public int compare(Object obj1, Object obj2) {

			return ((edge) obj1).cost - ((edge) obj2).cost;
		}

	}

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Hierarchy hr = new Hierarchy();
		Scanner Scan = new Scanner(System.in);
		int n = Scan.nextInt();
		int[] q = new int[n + 1];
		hr.Parent = new int[n + 1];
		hr.counter = new int[n + 1];
		hr.cnt = n;

		for (int i = 1; i <= n; i++)
			hr.Parent[i] = i;

		for (int i = 1; i <= n; i++)
			hr.counter[i] = 1;

		for (int i = 1; i <= n; i++)
			q[i] = Scan.nextInt();

		int m = Scan.nextInt();
		edge[] app = new edge[m];
		hr.visited = new boolean[n+1];

		for (int i = 0; i < m; i++) {
			Hierarchy.edge obj = hr.new edge();
			obj.supervisor = Scan.nextInt();
			obj.employee = Scan.nextInt();
			obj.cost = Scan.nextInt();
			app[i] = obj;

		}

		Hierarchy.Mycomp comp = hr.new Mycomp();
		Arrays.sort(app, 0, m, comp);

		
		
			for (int i = 0; i < m; i++) {
				if (!hr.visited[app[i].employee])
					if (q[app[i].supervisor] > q[app[i].employee])
						if (hr.find(app[i].supervisor) != hr.find(app[i].employee)) {
							hr.merge(app[i].supervisor, app[i].employee);
							hr.total += app[i].cost;
							hr.visited[app[i].employee]= true ;
							
						}
			}

			if(hr.cnt > 1)
				System.out.println("-1");
			else 
				System.out.println(hr.total);
		
	}

}
