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.*;


public class B
{
   public static void main(String[] args)
   {
      new B(new Scanner(System.in));
   }

   int[] qual;

   public B(Scanner in)
   {
      int N = in.nextInt();
      qual = new int[N];
      for (int i=0; i<N; i++)
         qual[i] = in.nextInt();
      
      int[] upTree = new int[N];
      Arrays.fill(upTree, -1);
      int[] cost = new int[N];
      Arrays.fill(cost, 987654321);
   
      int M = in.nextInt();
      for (int i=0; i<M; i++)
      {
         int p = in.nextInt()-1;
         int q = in.nextInt()-1;
         int w = in.nextInt();

         if (cost[q] > w)
         {
            cost[q] = w;
            upTree[q] = p;
         }
      }

      boolean passed = true;
      int cnt = 0;
      for (int i : upTree)
         if (i < 0)
            cnt++;

      if (cnt != 1)
         passed = false;

      int res = -1;
      if (passed)
      {
         res = 0;
         int i=0;
         for (int v : cost)
         {
            if (upTree[i] >= 0)
               res += v;
            i++;
         }
      }

      System.out.printf("%d%n", res);
   }
}
