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 PerfectPair
{
	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 <= 0 && y <= 0)
		{
			if(x >= m || y >= m)
			{
				System.out.println(0);
			}
			else
			{
				System.out.println("-1");
			}
		}
		else
		{
			long result = 0;
			
			if((x < m && y < m) && x + y <= 0)
			{
				result += -Math.min(x, y) / Math.max(x, y);
				
				if(x < y)
				{
					x += y * result;
				}
				else
				{
					y += x * result;
				}
			}
			
			while(x < m && y < m)
			{
				if(x <= y)
				{
					x += y;
				}
				else
				{
					y += x;
				}
				
				result++;
			}
			
			System.out.println(result);
		}
	}
}
