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

/**
 * Created by user on 28.06.2014.
 */
public class Main {

    public static void main(String[] args) throws IOException{

        Scanner sc = new Scanner(System.in);

        long a = sc.nextLong(),
             b = sc.nextLong(),
             m = sc.nextLong();

        long ans = 0;

        while ( true ){
            if ( a < b ) {
                a = a + b;
                b = a - b;
                a = a - b;
            }
            if ( a >= m ) break;
            if ( a <= 0 ){
                ans = -1;
                break;
            }
            if ( b >= 0 ){
                b += a;
                ++ ans;
                continue;
            } else {
                long temp = ( Math.abs( b ) + a - 1 ) / a;
                ans += temp;
                b += a * temp;
            }
        }

        System.out.println(ans);
    }
}
