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.math.*;
import java.util.*;
public class pp
{
public static void main(String ar[])
{
Scanner obj=new Scanner(System.in);
long x=obj.nextLong();
long y=obj.nextLong();
long m=obj.nextLong();
long count=0;
long min=Math.min(x,y);
long max=Math.max(x,y);
if(max>=m)
count=0;
if(min==0&&max==0&&m<=0)
count=0;
if(min==0&&max==0&&m>0)
count=-1;
if(min<0&&max<0&&m>max)
count=-1;
if(min<0&&max==0&&m>max)
count=-1;
if(min<0&&max>0&&max<m)
{
	count=((-min)/max)+1;
	min=min+(max*count);
}
min=Math.min(min,max);
max=Math.max(min,max);
if(min>=0&&max>0&&max<m)
{
	while(max<m&&min>=0)
	{
		long c=max;
		long d=max+min;
		min=Math.min(c,d);
		max=Math.max(c,d);
		++count;
	}
}
	System.out.println(count);
}
}
