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;

/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author Javi
 */
public class PerfectPair {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String[] input = in.nextLine().split(" ");
        long x = Long.parseLong(input[0]);
        long y = Long.parseLong(input[1]);
        long m = Long.parseLong(input[2]);

        long res = 0;
        if (x < m && y < m) {
            if (enoughToReachM(x, y)) {
                if(x < 0) {
                    res = (0-x)/y;
                    x = x + res*y;
                } else if (y < 0) {
                    res = (0-y)/x;
                    y = y + res*x;
                }
                while (m > x && m > y) {
                    if (x < y) {
                        x = x + y;
                    } else {
                        y = x + y;
                    }
                    res++;
                }
            } else {
                res = -1;
            }
        }

        System.out.println(res);

        in.close();
    }

    private static boolean enoughToReachM(long x, long y) {
        return x+y > x || x+y > y;
    }
}
