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 Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long x = sc.nextLong(), y = sc.nextLong(), m = sc.nextLong();

		if (Math.max(x, y) < m)
			if (Math.max(x, y) <= 0)
				System.out.println(-1);
			else {
				long c = 0;
				if (Math.min(x, y) < 0) {
					long min = Math.min(x, y);
					long max = Math.max(x, y);
					long k = (long) Math.ceil(Math.abs(min) * 1.0 / max);
					c = k;
					x = min + k * max;
					y = max;
//					System.out.println(x + " " + y + " k = " + k);
				}
				while (Math.max(x, y) < m) {
					if (x < y)
						x += y;
					else
						y += x;
					c++;

				}

				System.out.println(c);
			}
		else
			System.out.println(0);
	}

	/* 1000000000000000000 */
}
