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 static java.lang.System.*;
import static java.lang.Integer.*;

public class CF17B{
	public static void main(String[] args) throws Exception{
		BufferedReader sc = new BufferedReader(new InputStreamReader(in));
		
		int n = parseInt(sc.readLine());
		StringTokenizer st = new StringTokenizer(sc.readLine());
		
		ArrayList<Node> nodes = new ArrayList<Node>();
		for(int i=0; i<n; i++){
			nodes.add(new Node(parseInt(st.nextToken())));
		}
		
		int m = parseInt(sc.readLine());
		for(int i=0; i<m; i++){
			st = new StringTokenizer(sc.readLine());
			int a = parseInt(st.nextToken());
			int b = parseInt(st.nextToken());
			int c = parseInt(st.nextToken());
			if(nodes.get(b-1).best > c){
				nodes.get(b-1).best = c;
				nodes.get(b-1).par = nodes.get(a-1);
			}
		}
		
		Collections.sort(nodes);
		nodes.get(0).done = true;
		boolean ok = true;
		int ans = 0;
		for(int i=1; i<n; i++){
			Node no = nodes.get(i);
			if(no.best != MAX_VALUE && no.par.done){
				ans += no.best;
				no.done = true;
			}else{
				ok = false;
				break;
			}
		}
		
		if(ok){
			out.println(ans);
		}else{
			out.println(-1);
		}
	}
}

class Node implements Comparable<Node>{
	boolean done;
	Node par;
	int best;
	int q;
	public Node(int w){
		q = w;
		best = MAX_VALUE;
		done = false;
	}
	public int compareTo(Node n){
		return n.q-q;
	}
}