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.Scanner;

public class c_188 {
 public static void main(String[] args) {
  Scanner in = new Scanner(System.in);
  long x = in.nextLong();
  long y = in.nextLong();
  long m = in.nextLong();
  if (x >= m || y >= m) 
   System.out.println(0);
  else if (x <= 0 && y <= 0)
   System.out.println(-1);
  else {
   if (x > y) {
    long temp = x;
    x = y;
    y = temp;
   }
   long count = 0;
   if (x < 0) {
    count = -x / y;
    count++;
    x += y * count;
   }
   while (y < m) {
    long temp = x + y;
    x = y;
    y = temp;
    count++;
   }
   System.out.println(count);
  }
 }
}