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();
		long y = in.nextLong();
		long m = in.nextLong();
		in.close();				
		
		if (Math.max(x, y) >= m) {
			System.out.println(0);
		} else if (x + y <= Math.min(x, y)) {
			System.out.println(-1);
		} else {
			long sol = 0;
			
			if (x < 0) {
				long abs = x * -1;
				long initialIters = abs / y;
				sol = initialIters;
				x += initialIters * y;
			} else if (y < 0) {
				long abs = y * -1;
				long initialIters = abs / x;
				sol = initialIters;
				y += initialIters * x;
			}
			
			while (y < m && x < m) {
				sol++;
				if (x < y) {
					x += y;
				}
				else {
					y += x;
				}
			}
			System.out.println(sol);
		}
	}

}
