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.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {

	public static void main(String[] args) throws Exception {

		InputStreamReader in = new InputStreamReader(System.in);
		BufferedReader reader = new BufferedReader(in);

		String[] tokens = reader.readLine().split(" ");

		long x = Long.parseLong(tokens[0]);
		long y = Long.parseLong(tokens[1]);
		long m = Long.parseLong(tokens[2]);

		if (x >= m || y >= m) {
			System.out.println(0); // already m-perfect
			return;
		}

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

		long steps = 0;

		// make both numbers positive if one is negative
		if (x + y < 0) {
			long neg = Math.min(x, y);
			long pos = Math.max(x, y);
			x = pos;
			steps = (-neg) / pos;
			y = neg + steps * pos;
		}

		while (x + y < m) {
			if (x < y)
				x += y;
			else
				y += x;
			steps++;
		}
		System.out.println(steps + 1);

	}

}