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.InputStreamReader;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner r=new Scanner(new InputStreamReader(System.in));
        long x=r.nextLong(),y=r.nextLong(),m=r.nextLong(),t;
        if(x>y){t=x;x=y;y=t;}
        if(m<=y){System.out.print(0);return;}
        if(y<=0){System.out.print(-1);return;}
        long c=0;
        if(x<0){
            c=Math.abs(x)/y;
            x+=c*y;
        }
        while(m>y){
            c++;
            t=x+y;
            if(x<0)x=t;
            else{
                x=y;
                y=t;
            }
        }
        System.out.print(c);


    }
}
