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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/**
 * Created with IntelliJ IDEA.
 * User: shiwangi
 * Date: 6/14/13
 * Time: 9:32 PM
 * To change this template use File | Settings | File Templates.
 */
public class srm1883 {

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

            String tok[]=(br.readLine()).split(" ");
            long x=Long.parseLong(tok[0]);
            long y=Long.parseLong(tok[1]);
            long k=Long.parseLong(tok[2]);

            long count=0;int flag=0;

            if(Math.max(x,y)>=k) flag=1;
            else if(x<=0 && y<=0 ) flag=2;

            if(flag==0){
                if ( x< y && x<0) {
                    long absA = -x;
                    count+= absA/y;
                    x = -(absA % y);
                }
                if ( y< x && y<0) {
                    long absA = -y;
                    count+= absA/x;
                    y= -(absA % x);
                }

                    while(x+y<k)
                    {
                        if(x<y)
                            x=x+y;
                        else
                            y=x+y;
                        count++;
                    }


               }

            if(flag==1)
                System.out.println(0);
            else  if(flag==2)
                System.out.println(-1);
            else
                System.out.println(++count);
        }
    }