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.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
public class B {
	static boolean[][] edges;
	static int[] priority;
	public static void main(String[] args) throws IOException	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int v = Integer.parseInt(br.readLine());
		priority = new int[v];
		StringTokenizer st = new StringTokenizer(br.readLine());
		int e = Integer.parseInt(br.readLine());
		PriorityQueue<Edge> q = new PriorityQueue<Edge>();
		while(e-- > 0)	{
			st = new StringTokenizer(br.readLine());
			q.add(new Edge(Integer.parseInt(st.nextToken())-1, Integer.parseInt(st.nextToken())-1, Integer.parseInt(st.nextToken())));
		}
		boolean[] has = new boolean[v];
		long x = 0;
		while(!q.isEmpty())	{
			Edge ee = q.poll();
			if(!has[ee.b])	{
				has[ee.b] = true;
				x += ee.w;
			}
		}
		int num = 0;
		for(boolean b: has)
			if(!b)
				num++;
		System.out.println(num == 1 ? x : -1);
	}
	public static boolean isConnected(int a, int b, int l)	{
		boolean[] q = new boolean[l];
		LinkedList<Integer> queue = new LinkedList<Integer>();
		queue.add(a);
		while(!queue.isEmpty())	{
			int x = queue.removeFirst();
			if(q[x])
				continue;
			q[x] = true;
			if(x == b)
				return true;
			for(int i = 0; i < edges.length; i++)	{
				if(edges[x][i])	{
					queue.addLast(i);
				}
			}
		}
		return false;
	}
	static class Edge implements Comparable<Edge> {
		public int a,b,w;
		public Edge(int x, int y, int z)	{
			a=x;
			b=y;
			w=z;
		}
		public int compareTo(Edge e)	{
			return w-e.w;
		}
	}
}
