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;

/**
 * Created by mostafa on 9/9/17.
 */
public class PerfectPair {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long x = sc.nextLong(), y = sc.nextLong(), m = sc.nextLong();
        if(m > x && m > y && x <= 0 && y <= 0) {
            System.out.println(-1);
            return;
        }

        if(m <= x || m <= y) {
            System.out.println(0);
            return;
        }
        long steps = 0;
        long min = Math.min(x, y);
        long max = Math.max(x, y);
        while(m > max) {
            if(min < 0) {
                long nxt = (2 * max - min - 1) / (max);
                steps += nxt;
                min += nxt * max;
            }
            else {
                min += max;
                steps++;
            }

            long tmp = min;
            min = max;
            max = tmp;
        }

        System.out.println(steps);


    }
}
