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;

public class _317A {

	static long x, y, m, tmp, depth;
	static Scanner scanner = new Scanner(System.in);

	public static void main(String[] args) {
		depth = 0;
		x = scanner.nextLong();
		y = scanner.nextLong();
		m = scanner.nextLong();
		if (x < y) {
			tmp = x;
			x = y;
			y = tmp;
		}
		if (m > x) {
			if (x <= 0) {
				System.out.println("-1\n");
			}
			else {
				if (y < 0) {
					tmp = -y;
					tmp = tmp / x;
					y = y + x * (tmp+1);
					depth = (tmp+1);
				}
				int d = 0;
				do {
					tmp = y;
					y = x;
					x = x + tmp;
					d++;
				} while(x < m);
				System.out.println((depth + (long) d) + "\n");
				//cout << depth + (long long) d<< "\n";
				//1100087778366101931
			}
		}
		else {
//			cout << "0\n";
			System.out.println("0\n");
		}
	}

}
