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 Main {

	public static void main(String[] args) throws IOException{
		BufferedReader b = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(b.readLine());
		//don't actually need the q-values
		b.readLine();
		int[] q = new int[n];
		//init to most expensive
		for(int i=0; i<n; i++){
			q[i] = Integer.MAX_VALUE;
		}
		int m = Integer.parseInt(b.readLine());
		for(int i=0; i<m; i++){
			String[] si = b.readLine().split("\\s");
			int ai = Integer.parseInt(si[0]) - 1;
			int bi = Integer.parseInt(si[1]) - 1;
			int ci = Integer.parseInt(si[2]);
			//check value of boss, if cheaper, replace
			if(q[bi] > ci){
				q[bi] = ci;
			}
		}
		boolean possible = true;
		int sum = 0;
		for(int i=0; i<n; i++){
			if(q[i] == Integer.MAX_VALUE){
				if(!possible){
					System.out.println(-1);
					return;
				}
				possible = false;
				continue;
			}
			sum += q[i];
		}
		System.out.println(sum);
	}
}