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 C {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long x = sc.nextLong();
		long y = sc.nextLong();
		long m = sc.nextLong();

		if (x <= 0 && y <= 0 && m > 0)
			System.out.println(-1);
		// else if(x>=0 && y>=0 && m<)
		else if (x >= m || y >= m)
			System.out.println(0);
		else if(x<0 && y<0 &&(m>x && m>y))
			System.out.println(-1);
		else {
			long count=0;
			
			if(x<0)
			{
				long temp=x*-1;
				long t=temp%y ==0? (temp/y) :((temp/y)+1);
				count+=t;
				x+=t*y;
			}
			else if (y<0)
			{
				long temp=y*-1;
				long t=temp%x ==0? (temp/x) :((temp/x)+1);
				count+=t;
				y+=t*x;	
			}
			while(x<m && y<m)
			{
				count++;
				if(x>y)
				{
					y=y+x;
				}else
					x=x+y;
			}
			System.out.println(count);
//			
//			if (x>0 && y>0) {
//				long remx=m-y;
//				long devx=(remx%x==0)?(remx/x): (remx/x + 1);
//				long remy=m-x;
//				long devy=(remy%y==0)?(remy/y): (remy/y +1);
//				System.out.println(Math.min(devx, devy));
//			}
//			else if(x>0)
//			{
//				long remx=m-y;
//				long devx=(remx%x==0)?(remx/x): (remx/x+1);
//				System.out.println(devx);
//			}
//			else if (y>0)
//			{
//				long remy=m-x;
//				long devy=(remy%y==0)?(remy/y): (remy/y +1);
//				System.out.println(devy);
//			}
		}
	}

}
//-1000000000000000000 1 1000000000000000000