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.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;

public class A
{
	public static void main(String [] args) throws IOException
	{
		Scanner in = new Scanner(System.in);
		long x = in.nextLong();
		long y = in.nextLong();
		long m = in.nextLong();
		if(x > y)
		{
			long t = x;
			x = y;
			y = t;
		}

		if(y >= m)
		{
			System.out.println(0);
			return;
		}

		if(y <= 0)
		{
			System.out.println(-1);
			return;
		}

		long res = 0;
		long add = (Math.abs(x) + y - 1)/y;
		if(x < 0)
		{
			res = add;
			x += add*y;
			if(x > y)
			{
				long t = x;
				x = y;
				y = t;
			}
		}

		while(y < m)
		{
			x += y;

			if(x > y)
			{
				long t = x;
				x = y;
				y = t;
			}

			res++;

		}

		System.out.println(res);

	}
}