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 AperfectPair {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner input = new Scanner(System.in);
		long x = input.nextLong();
		long y = input.nextLong();
		long m = input.nextLong();
		if(Math.max(x,y)>0){
			long counter=0;
			if(y<x){
				long tmp = x;
				x = y;
				y = tmp;
			}
			if(m<=Math.max(x,y)){
				System.out.println(0);
				return;
			}
			if(x<0){
		    counter+=(Math.abs(x)+y-1)/y;
		    x+=counter*y;
		
			}
		    		while(Math.max(x,y)<m){
				if(x<y)x+=y;
				else
					y+=x;
				counter++;
		    		}
			System.out.println(counter);
		}
		else{
		   if(m<=Math.max(x,y)){	
			  System.out.println(0);
		   }
		   else
			   System.out.println(-1);
		}
	}
	

}
