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.*;

public class PerfectPair {

    public static void main (String []args) throws IOException {
    	BufferedReader br = new BufferedReader (new InputStreamReader (System.in));
    	String  str[] = br.readLine().split(" ");

    	long x = Long.parseLong(str[0]);
    	long y = Long.parseLong(str[1]);
    	long m = Long.parseLong(str[2]);

		System.out.println(count(x,y,m));
	}

	public static long count(long x,long y,long m){

		if((x<0  && y<0  && m==0) ||
		   (x<=0 && y<=0 && m>0 ) ||
		   (x<0  && y<0  && m<0 && m>x && m>y) )
			return -1;

		if(x>m || y>m)
			return 0;

		long count = 0l;
		long max = Math.max(x,y);
		long min = Math.min(x,y);

		if(min<0 && max>0){
			count = -1*min/max;
			min = min + count*max;
		}

		while(min<m && max<m){
			if(min<max)
				min = min+max;
			else
				max = min+max;

			count++;
		}

		return count;
	}
}