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.
//package projects;

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

public class CF17B
{
  public static void main(String[] args) throws IOException
  {
    BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
    int n = Integer.parseInt(sc.readLine());
    int[] qualifications = new int[n + 1];
    int total = 0;
    int max = 1000001;
    String[] lineItems = sc.readLine().split(" ");
    
    for (int i = 0; i <= n; i++)
    {
      qualifications[i] = max;
    }
    
    int m = Integer.parseInt(sc.readLine());
    
    for (int i = 0; i < m; i++)
    {
      lineItems = sc.readLine().split(" ");
      int a = Integer.parseInt(lineItems[0]);
      int b = Integer.parseInt(lineItems[1]);
      int c = Integer.parseInt(lineItems[2]);
      
      if (qualifications[b] > c)
      {
        qualifications[b] = c;
      }
    }
    
    int count = 0;
    for(int i = 1; i <= n; i++)
    {
      if(qualifications[i] == max)
      {
        count++;
      }
      else
      {
        total += qualifications[i];
      }
    }
    
    if(count > 1)
    {
      System.out.println("-1");
    }
    else
    {
      System.out.println(total);
    }
  }
}
