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 Igor
 */
public class C318 {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long x = in.nextLong();
        long y = in.nextLong();
        long m = in.nextLong();
        long res = 0;
        if (x <= 0 && y <= 0 && x < m && y < m) {
            System.out.println("-1");
            System.exit(0);
        }
		if (x >= m || y >= m) {
			System.out.println(0);
			System.exit(0);
		}
        if (x < 0 && y > 0) {
            long d = (Math.abs(x) / y);
            if (Math.abs(x) % y != 0) {
                d++;
            }
            res += d;
            x += d * y;
        } else if (x > 0 && y < 0) {          
            long d = Math.abs(y) / x;
            if (Math.abs(y) % x != 0) 
                d++;
            res += d;
            y += d * x;
        }
        while (true) {
            if (x >= m || y >= m) {
                break;
            }
            res++;
            if (x < y) {
                x = x + y;
            } else {
                y = x + y;
            }
        }
        System.out.println(res);
    }
}