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.math.BigInteger;
import java.util.*;

public class MPerfect{
	public static void main(String[] args){
		Scanner reader = new Scanner(System.in);
		long x = reader.nextLong();
		long y = reader.nextLong();
		long m = reader.nextLong();
		
		if(x >= m || y >= m){
			System.out.println(0);
			return;
		}
		if(x <= 0 && y <= 0){
			System.out.println(-1);
			return;
		} 
		
		BigInteger a = BigInteger.valueOf(Math.min(x,y));
		BigInteger b = BigInteger.valueOf(Math.max(x,y));
		BigInteger c = BigInteger.valueOf(m);
		
		long cnt = 0;
		
		if(a.compareTo(BigInteger.ZERO) < 0){
			BigInteger target = min(c,BigInteger.ZERO);
			BigInteger dist = target.subtract(a);
			BigInteger mult = dist.add(b.subtract(BigInteger.ONE)).divide(b);
			
			cnt += mult.longValue();
			BigInteger d = a.add(b.multiply(mult));
			
			a = min(b,d);
			b = max(b,d);
		}
		
		while(b.compareTo(c) < 0){
			BigInteger d = a.add(b);
			a = min(b,d);
			b = max(b,d);
			cnt++;
		}
		
		System.out.println(cnt);
	}
	
	public static BigInteger min(BigInteger a, BigInteger b){
		if(a.compareTo(b) < 0)
			return a;
		return b;
	}
	
	public static BigInteger max(BigInteger a, BigInteger b){
		if(a.compareTo(b) < 0)
			return b;
		return a;
	}
}
