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.*;
import java.io.*;
import java.math.*;
import javax.script.*;
import java.awt.*;

public class Hierarchy {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int size = sc.nextInt();
		int[] q=new int[size];
		for(int x=0;x<size;x++)
			q[x]=sc.nextInt();
		ArrayList<App> lis=new ArrayList<App>();
		int app=sc.nextInt();
		for(int x=0;x<app;x++)
		{
			int a=sc.nextInt()-1;
			int b=sc.nextInt()-1;
			int c=sc.nextInt();
			lis.add(new App(a,b, c));
		}
		Collections.sort(lis);
		int[] arr=new int[size];
		for(int x=0;x<size;x++)
			arr[x]=x;
		boolean[] done=new boolean[size];
		int cost=0;
		int need=0;
		for(int x=0;x<lis.size();x++)
		{
			if(Union(lis.get(x).a, lis.get(x).b, arr, done))
			{
				cost+=lis.get(x).c;
				need++;
			}
		}
		if(need+1==size)
		{
			System.out.println(cost);
		}
		else System.out.println(-1);
	}
	public static boolean Union(int a,int b,int[] arr,boolean[] done)
	{
		int pa=find(a,arr);
		int pb=find(b,arr);
		if(pa==pb || done[b])return false;
		arr[pa]=pb;
		done[b]=true;
		
		return true;
	}
	public static int find(int index,int[] arr)
	{
		if(arr[index]==index)return index;
		else return find(arr[index],arr);
	}
}
class App implements Comparable<App>
{
	int a;
	int b;
	int c;
	public App(int d,int e,int f)
	{
		a=d;
		b=e;
		c=f;
	}
	@Override
	public int compareTo(App o) {
		return c-o.c;
	}
	
}