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 in = new Scanner(System.in);
		long a = in.nextLong(), b = in.nextLong(), m = in.nextLong(), cnt = 0;
		while (true) {
			if (max(a, b) >= m) {
				System.out.println(cnt);
				break;
			} else if (max(a, b) <= 0) {
				System.out.println(-1);
				break;
			} else if (min(a, b) < 0 && max(a, b) > 0) {
				if (a < 0 && b > 0) {
					cnt += -(a / b) + 1;
					a += (-(a / b) + 1) * b;
				} else {
					cnt += -(b / a) + 1;
					b += (-(b / a) + 1) * a;
				}
			} else {
				cnt++;
				long c = a + b;
				if (a < b)
					a = c;
				else
					b = c;
				if (max(a, b) >= m) {
					System.out.println(cnt);
					break;
				}
			}
		}
	}

	public static long min(long a, long b) {
		return a > b ? b : a;
	}

	public static long max(long a, long b) {
		return a > b ? a : b;
	}
}

	 		 			 			   	 	 	  		    	 	