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 C {
	public static void main(String[] args) {
		Scanner myScanner = new Scanner(System.in);
		long a = myScanner.nextLong(), b = myScanner.nextLong(), m = myScanner
				.nextLong();

		if (a >= m || b >= m)
			System.out.println(0);
		else {
			if (a + b <= a && a + b <= b)
				System.out.println(-1);
			else {
				long cnt = 0;
				while (true) {
					if (a < b) {
						long tmp;
						if ((tmp = (Math.abs(a) / b)) > 1) {
							cnt += tmp;
							a += tmp * b;
						} else {
							a = a + b;
							cnt++;
						}
					} else {
						long tmp;
						if ((tmp = (Math.abs(b) / a)) > 1) {
							b += tmp * a;
							cnt += tmp;
						} else {
							b = a + b;
							cnt++;
						}
					}
					if (a >= m || b >= m) {
						System.out.println(cnt);
						break;
					}
				}
			}
		}
	}
}
