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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class MainClass {

	public static void main(String[] args) {
		int n = 0, i, m;

		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();

		DisjointSet ds = new DisjointSet(n);

		int[] qualifications = new int[n];

		for (i = 0; i < n; i++) {
			qualifications[i] = sc.nextInt();
			ds.makeSet(i);
		}

		m = sc.nextInt();
		ArrayList<Edge> applications = new ArrayList<Edge>();

		for (i = 0; i < m; i++) {
			int start, end, cost;
			start = sc.nextInt();
			end = sc.nextInt();
			cost = sc.nextInt();

			applications.add(new Edge(start, end, cost));
		}
		sc.close();

		Collections.sort(applications);

		int totalCost = 0;
		// for each edge in the set of edges
		for (Edge e : applications) {
			// if find-set u != find-set v
			if (ds.find(e.start - 1) != ds.find(e.end - 1) && !ds.hasParent(e.end - 1)) {
				totalCost += e.cost;
				ds.merge(e.start - 1, e.end - 1);
			}
		}

		// check that all nodes are connected
		int p = ds.find(0);
		boolean connected = true;
		for (i = 0; i < n; i++) {
			if (p != ds.find(i)) {
				connected = false;
				break;
			}
		}

		if (connected)
			System.out.println(totalCost);
		else
			System.out.println("-1");
	}
	
}

class DisjointSet {
	private int[] parents;
	private int[] rank;

	public DisjointSet(int size) {
		parents = new int[size];
		rank = new int[size];
	}

	public void makeSet(int element) {
		parents[element] = element;
		rank[element] = 0;
	}

	void merge(int first, int second) {
		int parentX = find(first);
		int parentY = find(second);

		if (rank[parentX] >= rank[parentY])
			parents[parentY] = parentX;
		else if (rank[parentX] < rank[parentY])
			parents[parentX] = parentY;

		if (rank[parentX] == rank[parentY])
			rank[parentY]++;
	}

	int find(int element) {
		if (parents[element] == element)
			return element;
		return find(parents[element]);
	}

	boolean hasParent(int element) {
		return parents[element] == element ? false : true;
	}

}


class Edge implements Comparable<Edge> {
	int start, end, cost;

	public Edge(int start, int end, int cost) {
		this.start = start;
		this.end = end;
		this.cost = cost;
	}

	public int getCost() {
		return this.cost;
	}

	@Override
	public int compareTo(Edge o) {
		if(this.cost > o.cost)
			return 1;
		else if(this.cost == o.cost)
			return 0;
		else
			return -1;
	}
}

