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 static java.lang.System.in;
import static java.lang.System.out;

import java.io.*;
import java.util.*;

public class C318 {
	static final double EPS = 1e-10;
	static final double INF = 1 << 31;
	static final double PI = Math.PI;

	public static Scanner sc = new Scanner(in);
	StringBuilder sb = new StringBuilder();
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

	public long run() throws IOException {
		String input;
		String[] inputArray;
		input = br.readLine();
		inputArray = input.split(" ");
		long x = Long.valueOf(inputArray[0]);
		long y = Long.valueOf(inputArray[1]);
		long m = Long.valueOf(inputArray[2]);
		long temp;
		if (x>y) { temp = x; x = y; y = temp; }
		if (y>=m) return 0;
		if (y<=0) return -1;
		long ans=0;
		if (x<0){
			long count =(-x)/y;
			ans+=count;
			x+=count*y;
			
		}
		do{
			temp = x;
			x = y;
			y = temp +y;
			if (x>y) { temp = x; x = y; y = temp; }
			ans++;
		}while (y<m);
		return ans;
	}

	public static void main(String[] args) throws IOException {
		long ans = new C318().run();
		ln(ans);
	}
	public static void ln(Object obj) {
		out.println(obj);
	}
}
