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.
//"We have two lives, and the second begins when we realize we only have one." — Confucius

import java.util.*;
import java.io.*;

public class A {
   public static void main(String[] args)
   {
	  FastScanner fs = new FastScanner();
	  long a= fs.nextLong(); long b= fs.nextLong(); long sum= fs.nextLong();
	  if((a<sum && b<sum) && (a<=0 && b<=0)) {System.out.println(-1);return;}
	  if(a>=sum || b>=sum) {
		  System.out.println(0);
		  return;
	  }
	  long count=0;
	  if(a<0 && b>0)
	  {
		  count = ((long)Math.abs(a)+(b-1))/b;
		  a=b*count+a;
	  }
	  else if(b<0 && a>0)
	  {
		  count=((long)Math.abs(b)+(a-1))/a;
		  b = a*count+b;
	  }
	  
	  
	  
	  while(a<sum && b<sum)
	  {
		  if(a<b)
		  {
			  a+=b;
			  count++;
		  }
		  else
		  {
			  b+=a;
			  count++;
		  }
	  }
	  System.out.println(count);
	  
	  
   }


	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}

	
}
