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 static java.lang.Math.*;
import java.util.*;

public class MPerfect {
	public static void main(String[] args) throws Exception {
		Scanner in = new Scanner(System.in);
		long x = in.nextLong();
		long y = in.nextLong();
		long m = in.nextLong();
		System.out.println(solve(x, y, m));
	}

	static long solve(long a, long b, long m) {
		long ans;
		if (a <= 0 && b <= 0 && m > a && m > b) {
			ans = -1;
		} else if (m < a || m < b) {
			ans = 0;
		} else {
			ans = 0;
			while (m > a && m > b) {
				long mn = min(a, b);
				long mx = max(a, b);
				if (mn < 0) {
					ans += (-mn / mx + 1);
					mn += (-mn / mx + 1) * mx;
				} else {
					mn += mx;
					ans += 1;
				}
				a = mn;
				b = mx;
			}
		}
		return ans;
	}
}
