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.*;
public	class	Main{
	static	long	OPCnt(long	a,long	b,long	m){
		long	s=0,t;
		if(a>b){t=a;a=b;b=t;}
		if(m<=b)	return	0;
		if(1>b)	return	-1;
		for(s=(b-a)/b,a+=b*s;m>b;++s){t=a+b;a=b;b=t;}
		return	s;
	}
	public	static	void	main(String[]args){
		Scanner	cin=new	Scanner(System.in);
		long	a=cin.nextLong(),b=cin.nextLong();
		System.out.print(OPCnt(a,b,cin.nextLong()));
	}
}
