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.Scanner;

public class C {

    /**
     * @param args
     */
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        long x = sc.nextLong();
        long y = sc.nextLong();
        long m = sc.nextLong();
        if(x >= m || y >= m){
            System.out.println(0);
            return;
        }

        if(x <= 0 && y <= 0){
            if(x >= m || y >= m){
                System.out.println(0);
            }
            else{
                System.out.println(-1);
            }
        }
        else if(x + y > 0){
            System.out.println(ope(x,y,m));
        }
        else{
            // make y min
            if(y > x){
                long tmp = x;
                x=y;y=tmp;
            }
            long yy = -y;
            long ans = 0;
            if(yy%x == 0){
                ans = yy/x;
                y = 0;
            }
            else{
                ans = yy/x + 1;
                y = x - yy%x;
            }
            System.out.println(ans + ope(x,y,m));
        }
    }
    static long ope(long x,long y,long m){
        long res = 0;
        while(x<m&&y<m){
            if(x < y){
                x += y;
            }
            else{
                y += x;
            }
            res++;
        }
        return res;
    }
}
