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.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
public class perfectPairr {
	
public static void main(String[] args) throws IOException{
			BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
			String x=bf.readLine();
			String [] y=x.split(" ");
			BigInteger n1=new BigInteger(y[0]);
			BigInteger n2=new BigInteger(y[1]);
			BigInteger n3=new BigInteger(y[2]);
			BigInteger counter=new BigInteger("0");
			BigInteger k=new BigInteger("1");
			BigInteger diff=new BigInteger("0");
			BigInteger diff2=new BigInteger("0");
			BigInteger big=new BigInteger("1000000");
			BigInteger div=new BigInteger("0");
			Boolean flag=true;
			BigInteger total=new BigInteger("0");
			if((n1.compareTo(n2)==-1))
			{
				diff=n3.subtract(n1);
				diff2=n2.subtract(n1);
			}
			else
			{
				diff=n3.subtract(n2);
				diff2=n1.subtract(n2);
				flag=false;
			}
			if(diff2.compareTo(big)==1)
			{
				if(flag)
				{
					div=n1.divide(n2);
					div=div.abs();
					total=div.multiply(n2);
					n1=n1.add(total);
					counter=counter.add(div);
				}
				else
				{
					div=n2.divide(n1);
					div=div.abs();
					total=div.multiply(n1);
					n2=n2.add(total);
					counter=counter.add(div);
				}
			}
			if((n1.compareTo(n3)==0)||(n1.compareTo(n3)==1)||(n2.compareTo(n3)==0)||(n2.compareTo(n3)==1))
			{
				System.out.println(0);
				return;
			}
			while((n1.compareTo(n3)==-1)&&(n2.compareTo(n3)==-1))
			{
				if((n1.compareTo(n2)==-1))
				{
					n1=n1.add(n2);
					diff2=n3.subtract(n1);
					if((diff2.compareTo(diff)==0)||(diff2.compareTo(diff)==1))
					{
						System.out.println(-1);
						return;
					}
					counter=counter.add(k);
				}
				else
				{
					n2=n2.add(n1);
					diff2=n3.subtract(n2);
					if((diff2.compareTo(diff)==0)||(diff2.compareTo(diff)==1))
					{
						System.out.println(-1);
						return;
					}
					counter=counter.add(k);
				}
			}
			System.out.println(counter);

}
}
