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 perfect {

    public static void main (String[]args) {
    	Scanner sc=new Scanner (System.in);
    	long x=sc.nextLong();
    	long y = sc.nextLong();
    	long m=sc.nextLong();
    	long n=0;
    	while (true){
    		if(x>=m|y>=m){
    			System.out.println(n);
    			break;}
    		else
    			if(x<=0&&y<=0){
    				System.out.println("-1");
    				break;}
    			else{
    				if(x>y)
    				  	if(y<0&x>0)
    				  	{ n=-1*y/x+1;
    				  	  y=x-(-1*y)%x;
    				  	}
    				  	else
    				  	{	y=x+y;
    				  	n++;}
    				else
    				  	if(x<0&y>0)
    				  	{ n=-1*x/y+1;
    				  	  x=y-(-1*x)%y;
    				  	}
    				    else
    				    {x=x+y;
    				    n++;}
    				}

    			}
    }


}