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.*;

public class IerarhiyaNew implements Runnable{
	public static void main(String[] args) throws IOException {
		new Thread(new IerarhiyaNew()).start();
	}
	
	BufferedReader br;
	StringTokenizer in;
	PrintWriter out;
	
	public String nextToken() throws IOException{
		while (in == null || !in.hasMoreTokens()){
			in = new StringTokenizer(br.readLine());
		}
		
		return in.nextToken();
	}
	
	public int nextInt() throws IOException{
		return Integer.parseInt(nextToken());
	}
	
	public double nextDouble() throws IOException{
		return Double.parseDouble(nextToken());
	}
	
	public class pair implements Comparable<pair>{
		int q, i;
		
		public int compareTo(pair o){
			if (q < o.q)return 1;
			if (q > o.q)return -1;
			return 0;
		}
	}
	
	public void solve() throws IOException{
		int n = nextInt();
		
		int[] q = new int[n];
		
		for (int i = 0; i < n; i++) {
			q[i] = nextInt();
		}
		
		int m = nextInt();
		
		int[] min = new int[10000];
		
		Arrays.fill(min, Integer.MAX_VALUE);
		
		for (int i = 0; i < m; i++) {
			int a = nextInt() - 1;
			int b = nextInt() - 1;
			int c = nextInt();
			
			if (min[b] > c)min[b] = c;
		}
		
		int ans = 0;
		int f = 0;
		
		for (int i = 0; i < n; i++) {
			if (min[i] == Integer.MAX_VALUE){
				f++;
				if (f == 2){
					ans = -1;
					break;
				}
			}
			else
				ans += min[i];
		}
		
		out.println(ans);
	}
	
	public void run(){
		try{
			br = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			
			solve();
			
			out.close();
		}
		catch(IOException e){
			e.printStackTrace();
			System.exit(1);
		}
	}
}
