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.Scanner;

public class C {
	public static void main(String [] args){
		Scanner in = new Scanner(System.in);
		long x = in.nextLong();
		long y = in.nextLong();
		long m = in.nextLong();
		if(y >= m || x>=m){
			System.out.println(0);
			return;
		}
		if(x == 0 && y == 0){
			System.out.println(-1);
			return;
		}
		if(x * y == 0 ){
			if(x*m <0 || m*y <0){
				System.out.println(-1);
				return;
			}
			else{
				x = Math.abs(x);
				y = Math.abs(y);
				m = Math.abs(m);
				long k = Math.max(x, y);
				y = k;
				x = 0;
				long p = 0;
				long n = 1;
				long c = 0;
				while(n*y<m){
					long t = p;
					p = n;
					n += t;
					c++;
				}
				System.out.println(c);
				return;
			}
		}
		if(x*y > 0 && ( (y>0 && m <=0) || (y<0 && m>=0 ) ) ){
			System.out.println(-1);
			return;
		}
		if(Math.abs(m-y) > Math.abs(m-x)){
			long t = y;
			y = x;
			x = t;
		}
		if(Math.abs(x)>Math.abs(m) && m*x >0){
			System.out.println(-1);
			return;
		}
		long c = 0;
		if(x*y<0){
			if(y<0){
				y*=-1;
				m*=-1;
			}
			else{
				x*=-1;
			}
			c+=x/y;
			x = -1*x + c*y;
			if(x<0){
				c++;
				x += y;
			}
			
		}
		long p = 0;
		long n = 1;
		long k = n*y+p*x;
		while(k<m){
			long t = p;
			p = n;
			n += t;
			k = n*y+p*x;
			c++;
		}
		System.out.println(c);
		
	}
}
