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.
import java.util.*;

public class Main{
   
  static long max(long x , long y){
      return x > y ? x : y;
  }
  
  static long solve(long a ,long b ,long c){
      long res = 0;
      while( (a >= c || b >= c) == false){
          long sum = a + b;
          long Max = max(a , b);
          a = sum; b = Max;
          res++;
      }
      return res;
  }
  static long __solve(long a , long b , long c){
      long res = (-b + a - 1) / a;
      long tmp = b + res * a;
      res += solve(a,tmp,c);
      return res;
  }
  
  public static void main(String[] args){
      Scanner cin = new Scanner(System.in);
      long a,b,c,res;
      a = cin.nextLong();
      b = cin.nextLong();
      c = cin.nextLong();
      if(a >= c || b >= c)
          res = 0;
      else if(a <= 0 && b <= 0)
          res = -1;
      else if(a >= 0 && b >= 0){
          res = solve(a,b,c);
      }
      else{
          long tmp;
          if(a < b){
              tmp = a; a = b; b = tmp;
          }
          res = __solve(a,b,c);
      }
      
      System.out.println(res);
  }
}