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.*;
import java.io.*;

public class third
{
	public static void main(String args[])
	{
		long n,m,k;
		Scanner sc = new Scanner(System.in);
		n = sc.nextLong();
		m = sc.nextLong();
		k = sc.nextLong();
		if(n>=k || m>=k)
		System.out.println(0);
		else
		{
			if(n<=0 && m<=0 )
			{
				System.out.println(-1);
			}
			else
			{
				long x=Math.max(n,m);
				long y=Math.min(n,m);
				long iter = 0;
				if(y<0)
				{
					long div = (Math.abs(y))/x;
					long rem = Math.abs(y)%x;
					iter=div;
					y=(-1)*rem;
					if(rem>0){
						iter++;
						y+=x;
					}
					
				}
				int i=0;
				while(x<k)
				{
					long temp = x;
					x=x+y;
					y=temp;
					i++;
				}
				System.out.println(iter+i);
			}
		}
		
	}
}
