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.util.Scanner;

public class C {
    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner(System.in);
        long x = in.nextLong(), y = in.nextLong(), m = in.nextLong();

        if (m <= x || m <= y)
            System.out.println(0);
        else {
            if (x <= 0 && y <= 0) {
                System.out.println(-1);
            } else {
                long count = 0;
                while(x < 0 || y < 0){
                    long min = Math.min(x, y), max = Math.max(x, y);
                    long z = (long) Math.ceil(Math.abs(min * 1.0 / max));
                    x = max;
                    y = min + z * max;
                    count += z;
                }
                
                while (x < m && y < m) {
                    long min = Math.min(x, y), max = Math.max(x, y);
                    x = max;
                    y = min + max;
                    count++;
                }

                System.out.println(count);
            }
        }
    }
}
