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 PerfectPair {
	public static void main(String[] args){
		Scanner in=new Scanner(System.in);
		long x=in.nextLong(),y=in.nextLong(), m=in.nextLong();
		if(m<=x || m<=y){
			System.out.println(0);
		}
		else if((x<=0 && y<=0))
			System.out.println(-1);
		else{
			long add=0;
			long max=Math.max(x, y);
			long min=Math.min(x, y);
			x=max;
			y=min;
			if(x<0 || y<0){
				long div=min/max;
				y=min-div*max + max;
				add=div*(-1)+1;
				
			}
			long i=0;
			while(x<m){
				long change=x;
				x=x+y;
				y=change;
				i++;
			}
			System.out.println(add+i);
		}	
	}
}
