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.HashMap;
import java.util.Scanner;


public class PerfectPair {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		long a = in.nextLong();
		long b = in.nextLong();
		long m = in.nextLong();
		if (a>b) {
			b = a+b;
			a = b-a;
			b = b-a;
		}
		if (b<m && a+b <= a) {
			System.out.println("-1");
			return;
		}
		long tmp,ans =0;
		
		if (b >= m) {
			System.out.println(0);
			return;
		}
		if (a < 0 && -a > b) {
			ans = -a / b + 1;
			a = a + ans * b;
		}
		
		while (b<m){
			tmp = b;
			b = b+a;
			a = tmp;
			
			ans++;
			
			if (a>b) {
				b = a+b;
				a = b-a;
				b = b-a;
			}
		}
		System.out.println(ans);
	}

}
