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.
/**
 * Created with IntelliJ IDEA.
 * User: yuantian
 * Date: 6/16/13
 * Time: 10:19 PM
 * Copyright (c) 2013 All Right Reserved, http://github.com/tyuan73
 */

import java.util.Scanner;

public class PerfectPair317A {
    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 (y < x) {
            long t = x;
            x = y;
            y = t;
        }

        if (y >= m) {
            System.out.println(0);
            return;
        }
        if (y <= 0) {
            System.out.println(-1);
            return;
        }

        // y > 0 and y < m
        long total = 0;
        while (y < m) {
            if (x < 0) {
                if (x % y == 0) {
                    total += (y - x) / y;
                    x = y;
                } else {
                    total += (y - x) / y + 1;
                    long t = y * total + x;
                    x = y;
                    y = t;
                }
            } else {
                long t = x + y;
                x = y;
                y = t;
                total++;
            }
        }
        System.out.println(total);
    }
}
