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 B
{
	public static int[] p;

	public static void main(String args[])
	{
		Scanner scan = new Scanner(System.in);

		int n = scan.nextInt();

		int[] q = new int[n];
		for(int i=0;i < n;i++)
			q[i] = scan.nextInt();

		int m = scan.nextInt();

		Edge[] e = new Edge[m];

		for(int i=0;i < m;i++)
			e[i] = new Edge(scan.nextInt()-1, scan.nextInt()-1, scan.nextInt());

		p = new int[n];

		for(int i=0;i < n;i++)
			p[i] = i;


		Arrays.sort(e);

		int rtn = 0;

		for(int i=0;i < m;i++)
		{
			if(find(e[i].u) != find(e[i].v) && find(e[i].v) == e[i].v)
			{
				union(e[i].u, e[i].v);
				rtn += e[i].c;
			}
		}

		int boss = find(p[0]);
		for(int i=0;i < n;i++)
		{
			if(find(p[i]) != boss)
				rtn = -1;
		}

		System.out.println(rtn);

	}


	public static int find(int x)
	{
		if(p[x] == x)
			return x;

		int root = find(p[x]);
		p[x] = root;

		return root;
	}

	public static void union(int x, int y)
	{
		p[y] = find(x);
	}
}


class Edge implements Comparable
{
	int u, v, c;

	public Edge(int iu, int iv, int ic)
	{
		u = iu; v = iv; c = ic;
	}

	public int compareTo(Object o)
	{
		Edge that = (Edge)o;

		return this.c - that.c;
	}
}