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

public class A {
	public static void main(String[] args) throws IOException {
		st = new StringTokenizer(rd.readLine());
		x = Long.parseLong(st.nextToken());
		y = Long.parseLong(st.nextToken());
		m = Long.parseLong(st.nextToken());
		if(x>=m || y>=m){
			System.out.println(0);
			return;
		}
		if(x<=0 && y<=0){
			System.out.println(-1);
			return;
		}
		if(x>=0 && y>=0){
			int num = 0;
			while(true){
				if(x>=m || y>=m) break;
				long max = Math.max(x, y);
				long sum = x+y;
				x = max;
				y = sum;
				num++;
			}
			System.out.println(num);
		}
		else{
			long min = Math.min(x, y);
			long other = x+y-min;
			x = min;
			y = other;
			// x+ky>=0
			long k = (-x+y-1)/y;
			x = x+k*y;
			long num = k;
			while(true){
				if(x>=m || y>=m) break;
				long max = Math.max(x, y);
				long sum = x+y;
				x = max;
				y = sum;
				num++;
			}
			System.out.println(num);
		}
	}
	

	
	static long x, y, m;
	
	static StringTokenizer st;
	static BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter pw = new PrintWriter(System.out);
	
}
