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

public class Main {
	static BufferedReader br = new BufferedReader( new InputStreamReader(System.in) );
	static StringTokenizer st = new StringTokenizer("");
	static String next()	throws Exception	{
		while	( !st.hasMoreTokens() )	st = new StringTokenizer( br.readLine() );
		return	st.nextToken();
	}
	public static void main(String [] asda)	throws Exception	{
		PrintWriter out = new PrintWriter( new BufferedOutputStream(System.out) );
		//
		long x = Long.parseLong( next() );
		long y = Long.parseLong( next() );
		long M = Long.parseLong( next() );
		long aux, moves;
		if	( Math.max(x, y) <= 0 )	out.println( Math.max(x, y) >= M ? 0 : -1 );
		else if	( Math.max(x, y) >= M )	out.println(0);
		else	{
			moves = 0;
			if	( Math.min(x, y) < 0 )	{
				moves = Math.abs( Math.min(x, y) ) / Math.max(x, y);
				aux = moves * Math.max(x, y);
				if ( x < y )	x += aux;
				else	y += aux;
			}
			while	( Math.max(x, y) < M )	{
				aux = Math.min(x, y) + Math.max(x, y);
				y = Math.max(x, y);
				x = aux;
				moves++;
			}
			out.println(moves);
		}
		out.flush();
	}
}

