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.StringTokenizer;

public class C {

	public static void main(String[] args) throws IOException {
		PrintWriter out = new PrintWriter(System.out);
		sc = new StringTokenizer(br.readLine());
		long x = nxtLng();
		long y = nxtLng();
		long m = nxtLng();
		if (x >= m || y >= m)
			out.println(0);
		else {
			if (x <= 0 && y <= 0)
				out.println(-1);
			else {
				if (x >= 0 && y >= 0) {
					long res = 0;
					while (x < m && y < m) {
						long num = x + y;
						if (x < y)
							x = num;
						else
							y = num;
						res++;
					}
					out.println(res);
				} else if (x < 0) {
					long tmp = -1 * x;
					long res = tmp / y;
					res++;
					x = y - (tmp % y);
					while (x < m && y < m) {
						long num = x + y;
						if (x < y)
							x = num;
						else
							y = num;
						res++;
					}
					out.println(res);
				} else {
					long tmp = -1 * y;
					long res = tmp / x;
					res++;
					y = x - (tmp % x);
					while (x < m && y < m) {
						long num = x + y;
						if (x < y)
							x = num;
						else
							y = num;
						res++;
					}
					out.println(res);

				}
			}
		}
		br.close();
		out.close();
	}

	static BufferedReader br = new BufferedReader(new InputStreamReader(
			System.in));

	static StringTokenizer sc;

	static String nxtTok() throws IOException {
		while (!sc.hasMoreTokens()) {
			String s = br.readLine();
			if (s == null)
				return null;
			sc = new StringTokenizer(s.trim());
		}
		return sc.nextToken();
	}

	static int nxtInt() throws IOException {
		return Integer.parseInt(nxtTok());
	}

	static long nxtLng() throws IOException {
		return Long.parseLong(nxtTok());
	}

	static double nxtDbl() throws IOException {
		return Double.parseDouble(nxtTok());
	}

	static int[] nxtIntArr(int n) throws IOException {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nxtInt();
		return a;
	}

	static long[] nxtLngArr(int n) throws IOException {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nxtLng();
		return a;
	}

	static char[] nxtCharArr() throws IOException {
		return nxtTok().toCharArray();
	}
}