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 A {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		long a=sc.nextLong();
		long b=sc.nextLong();
		long m=sc.nextLong();
		if(a>=m||b>=m) {
			System.out.println(0);
		}else if(a<=0&&b<=0){
			System.out.println(-1);
		}else{
			long res=0;
			if(a<0){
				res+=Math.ceil(Math.abs(a)/(b*1.0));
				a+=res*b;
			}else if(b<0){
				res+=Math.ceil(Math.abs(b)/(a*1.0));
				b+=res*a;
			}
			long max=Math.max(a,b);
			long min=Math.min(a,b);
			while(max<m){
				res++;
				long tmp=max;
				max+=min;
				min=tmp;
			}
			System.out.println(res);
		}
	}
}
