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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;


public class Main {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        
        int employeeCount = Integer.parseInt(br.readLine());
        br.readLine();
        int applicationCount = Integer.parseInt(br.readLine());
        
        int[] costArray = new int[employeeCount];
        for (int i = 0; i < employeeCount; i++) {
            costArray[i] = -1;
        }
        
        for (int i = 0; i < applicationCount; i++) {
            String[] application = br.readLine().split(" ");
            int a = Integer.parseInt(application[0]), b = Integer.parseInt(application[1]), c = Integer.parseInt(application[2]);
            // Get to zero based indices
            a--; b--;
            
            if (costArray[b] == -1) {
                costArray[b] = c;
            } else {
                costArray[b] = Math.min(costArray[b], c);
            }
            
        }
        
        int count = 0;
        int sum = 0;
        for (int i = 0; i < employeeCount; i++) {
            
            sum+=costArray[i];
            
            if (costArray[i] == -1) {
                count++;
            }
            
            if (count > 1) {
                System.out.println("-1");
                break;
            }
        }
        
        if (count <= 1) {
            System.out.println(sum+1);
        }
        
        
    }

}
