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.
/**
 * Created with IntelliJ IDEA.
 * User: NurgalievAT
 * Date: 14.06.13
 * Time: 19:36
 * To change this template use File | Settings | File Templates.
 */

import java.util.*;
import java.math.*;

public class Mian {
    public static void main(String[] args)
    {
        Scanner in = new Scanner(System.in);
        BigInteger x = in.nextBigInteger(), y = in.nextBigInteger(), m = in.nextBigInteger();

        if(x.compareTo(m)>=0 ||y.compareTo(m)>=0)
        {
            System.out.println(0);
            return;
        }
        if(x.compareTo(y)>0)
        {
            BigInteger t = x;
            x = y;
            y = t;
        }
        BigInteger ans = BigInteger.ZERO;
        if(x.compareTo(BigInteger.ZERO)<0 && y.compareTo(BigInteger.ZERO)>0)
        {
            BigInteger les = x;
//            if(m.compareTo(BigInteger.ZERO)<0)
//                les = m;
            les = les.abs();
            BigInteger d = les.divide(y);
            ans = d;
            x = x.add(d.multiply(y));
        }
        for(int i=0;i<200;i++)
        {
            if(x.compareTo(m)>=0 || y.compareTo(m)>=0)
            {
                System.out.println(ans.add(BigInteger.valueOf(i)));
                return;
            }
            if(x.compareTo(y)<0)
            {
                x = x.add(y);
            }else{
                y = y.add(x);
            }
        }
        System.out.println(-1);
    }

}
