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.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class c_188 {
	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 >= m || y >= m)
			System.out.println(0);
		else if (x <= 0 && y <= 0)
			System.out.println(-1);
		else {
			if (x > y) {
				long temp = x;
				x = y;
				y = temp;
			}
			long count = 0;
			if (x < 0) {
				count = -x / y;
				count++;
				x += y * count;
			}
		//	System.out.println(x+" "+y+" "+count);
			while (y < m) {
				long temp = x + y;
				x = y;
				y = temp;
				count++;
			}
			System.out.println(count);
		}
	}
}
