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.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class Acm {
	public static void main(String args[]){ 
		Scanner sc= new Scanner(System.in);long c= 0 ;
		long x =sc.nextLong(); long y= sc.nextLong(); 
		long m = sc.nextLong();
		if(x>=m || y>=m) System.out.println(0);
		else{
			 if(x<=0 && y<=0  && x<=m && y<=m) 
				System.out.println(-1);
			else{
				long min =Math.min(x, y);
                if(x<0){ c = -x/y   ;  x +=c*y;
				//
				}
                else if(y<0){ c = -y/x   ;  y +=c*x;
				
				}
                // now x and y positive grow exponentially :) 
				while(true){
					if(x>=m || y>=m){
						System.out.println(c); break ;
					}
					if(x<y) x+=y ;
					else y+=x;
					
				
				
				c++;
				
				
					
				 
					
				}
				
				
				
				
				
			}
			
			
			
		}
			
			
		}
			
			
		

		
		
		
		

			
				
	
		
			
			
			
			
	}