Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.

Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).

What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?

Input

Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.

Output

Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.

Examples

Input

1 2 5


Output

2


Input

-1 4 15


Output

4


Input

0 -1 5


Output

-1

Note

In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).

In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).

Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
/*
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 * and open the template in the editor.
 */


import java.util.*;
import java.lang.*;
/**
 *
 * @author User
 */
public class Con188A {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        Scanner sc = new Scanner (System.in);
        long a = sc.nextLong();
        long b = sc.nextLong();
        long m = sc.nextLong();
        boolean returnFlag = ((a<=0)&&(b<=0)&&(Math.max(a, b)<m))? false:true;
        long returnCode = returnFlag ? 0:-1;
        if ((Math.min(a, b)<0)&&(Math.max(a, b)>0)&&(Math.max(a, b)<m))
                  {
                     long returnIncrement = (a>b) ? -b/a : -a/b;
            
                     returnCode+=returnIncrement;
                      if (a>b) {
                          b = b + a * returnIncrement;
                      }
                      else {
                          a = a + b * returnIncrement;
                      }
                  }
                  
        while (returnFlag)
          {
              if (Math.max(a, b)<m)
              {
                               
                  if (a>b) b+=a;
                  else a+=b;
                  returnCode++;
              }
              else returnFlag=false;
          };
        System.out.print(returnCode);
    }
}
