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.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Locale;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class Solution {
	
	BufferedReader in;
	PrintWriter out;
	StringTokenizer st;
	
	String nextToken() throws Exception {
		if (st == null || !st.hasMoreTokens()) st = new StringTokenizer(in.readLine());
		return st.nextToken();
	}
	
	int nextInt() throws Exception {
		return Integer.parseInt(nextToken());
	}
	
	long nextLong() throws Exception {
		return Long.parseLong(nextToken());
	}
	
	void solve() throws Exception {
		long x = nextLong();
		long y = nextLong();
		long m = nextLong();
		if (x > y) { long t = x; x = y; y = t; }
		if (y < m && m <= 0 || m > 0 && y <= 0) {
			out.println(-1);
		} else {
			long ans = 0;
			while (y < m) {
				if (x < 0) {
					long t = Math.abs(x) / y + 1;
					x += y * t;
					ans += t;
				} else {
					x += y;
					if (x > y) { long t = x; x = y; y = t; }
					ans++;
				}
			}
			out.println(ans);
		}
	}
	
	void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			Locale.setDefault(Locale.US);
			solve();
		} catch(Exception e) {
			e.printStackTrace();
			System.exit(1);
		} finally {
			out.close();
		}
	}

	public static void main(String[] args) {
		new Solution().run();
	}
	
}