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;

/**
 *
 * @author 743012015
 */
public class CR188C {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        long x = in.nextLong();
        long y = in.nextLong();
        long m = in.nextLong();

        if (Math.max(x, y) >= m) {
            System.out.println("0");
            return;
        }

        if (x <= 0 && y <= 0) {
            System.out.println("-1");
            return;
        }

        long r = 0;

        if (x < 0) {
            r = (Math.abs(x) + y) / y;
            x += r * y;
        }

        if (y < 0) {
            r = (Math.abs(y) + x) / x;
            y += r * x;
        }

        while (Math.max(x, y) < m) {
            long s = x + y;
            if (x <= y) {
                x = s;
            } else {
                y = s;
            }
            ++r;
        }

        System.out.println(r);
    }
}
