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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;

public class A {
	public static BufferedReader in;
	public static PrintWriter out;

	public static void main(String[] args) throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);

		boolean showLineError = true;
		if (showLineError) {
			solve();
			out.close();
		} else {
			try {
				solve();
			} catch (Exception e) {
			} finally {
				out.close();
			}
		}

	}

	static void debug(Object... os) {
		out.println(Arrays.deepToString(os));
	}

	private static void solve() throws IOException {
		String[] line = in.readLine().split(" ");
		long x = Long.parseLong(line[0]);
		long y = Long.parseLong(line[1]);
		long m = Long.parseLong(line[2]);

		long[] arr = { x, y };
		Arrays.sort(arr);
		if (arr[1] >= m) {
			out.println(0);
			return;
		}
		if (arr[1] <= 0L) {
			out.println(-1);
			return;
		}

		long ret = 0;
		if (arr[0] < 0L) {
			long t = -arr[0];
			ret = (t + arr[1] - 1L) / arr[1];
			arr[0] += ret * arr[1];
		}
		if (arr[0] == 0L) {
			ret++;
			arr[0] += arr[1];
		}
		out.println(ret + go(arr, m));

	}

	private static long go(long[] arr, long m) {
		Arrays.sort(arr);
		if (arr[1] >= m)
			return 0L;
		long ret = log(arr[0], arr[1], m);
		return Math.min(ret, 1L + go(new long[] { arr[1], arr[0] + arr[1] }, m));

	}

	private static long log(long a, long b, long m) {
		return (m - b + a - 1L) / a;
	}

}