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.*;
import java.util.*;

public class Main {

	public static void main(String[] args) throws Exception {
		long time = System.nanoTime();

		long x = nextl();
		long y = nextl();
		long m = nextl();

		if (x >= m || y >= m) out.println(0);
		else {
			if (x <= 0 && y <= 0) out.println(-1);
			else {
				long c = 0;
				if (x < 0) {
					c += (y - x)/y;
					x += c*y;
				} else if (y < 0) {
					c += (x - y)/x;
					y += c * x;
				}

				while (x < m && y < m) {
					if (x > y) y += x;
					else x += y;
					c++;
				}
				out.println(c);
			}
		}
		
//		System.out.println((System.nanoTime() - time)* 1e-9);
		out.close();
	}
	static PrintWriter out = new PrintWriter(System.out);
	
	static BufferedReader bufferedreader = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer in = new StringTokenizer("");

	static String nextToken() throws Exception {
		if (!in.hasMoreTokens()) in = new StringTokenizer(bufferedreader.readLine());
		return in.nextToken();
	}

	static int next()  throws Exception {return Integer.parseInt(nextToken());};
	static int[] next(int n) throws Exception {
		int[] x = new int[n];
		for (int i = 0; i < n; i++) x[i] = next();
		return x;
	}
	static int[][] next(int n, int m) throws Exception {
		int[][] x = new int[n][];
		for (int i = 0; i < n; i++) x[i] = next(m);
		return x;
	}

	static long nextl() throws Exception {return Long.parseLong(nextToken());};
	static long[] nextl(int n) throws Exception {
		long[] x = new long[n];
		for (int i = 0; i < n; i++) x[i] = nextl();
		return x;
	}
	static long[][] nextl(int n, int m) throws Exception {
		long[][] x = new long[n][];
		for (int i = 0; i < n; i++) x[i] = nextl(m);
		return x;
	}

	static double nextd() throws Exception {return Double.parseDouble(nextToken());};
	static double[] nextd(int n) throws Exception {
		double[] x = new double[n];
		for (int i = 0; i < n; i++) x[i] = nextd();
		return x;
	}
	static double[][] nextd(int n, int m) throws Exception {
		double[][] x = new double[n][];
		for (int i = 0; i < n; i++) x[i] = nextd(m);
		return x;
	}

	static String nextline() throws Exception {
		in = new StringTokenizer("");
		return bufferedreader.readLine();
	}

}