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;

public class Main {
	static long count = 0;

	public static void main(String[] args) throws IOException {
		BufferedReader br1 = new BufferedReader(
				new InputStreamReader(System.in));
		String data[] = br1.readLine().split(" ");
		long x = Long.parseLong(data[0]);
		long y = Long.parseLong(data[1]);
		long m = Long.parseLong(data[2]);
		if(x>=m || y>=m)
			System.out.println(0);
		else if(m < 0)
		{
			if(x>=m)
			{
				System.out.println(0);
			}
			else if(y>=m)
			{
				System.out.println(0);
			}
			else{
				System.out.println(-1);
			}
		}
		else if(m>0)
		{
			if(x>0 || y>0)
			{
				
				if(x+y>=m)
				{
					System.out.println(1);
					return;
				}
				else if(x>m || y>m)
				{
					System.out.println(0);
					return;
				}
				else 
				{
					if(x>0 && y >0)
					{
						dp(x,y,m);
						System.out.println(count);
					}
					else
					{
						//System.out.println("came here");
						long neg = x>0 ? y : x;
						long d = x>0 ? x :y;
						long c = (long)Math.ceil(Math.abs(neg)/d);
						count+= c;
						dp(d,d*c+neg,m);
						System.out.println(count);
						
					}
					
				}
				
			}
			else 
			{
				System.out.println(-1);
				
			}
				
			return;
				
		}
		else if(m==0)
		{
			if(x==0 || y==0)
				System.out.println(0);
			else if(x+y>=0)
				System.out.println(0);
			else
				System.out.println(-1);
		}
		

	}

	static void dp(long x, long y, long m) {
		// count++;
		// System.out.println(x+" "+y+" ");
		//System.out.println(x + " " + y);
		if (x + y >= m) {
			count++;
			return;
		} else {
			count++;
			if (x <= y) {
				dp(x + y, y, m);
			}

			else {
				dp(x, x + y, m);
			}
		}

	}
	
	static void dps(long x,long y,long m)
	{
		while(true)
		{
			if(x+y>=m)
				{count++;
				break;
				}
			if(x>=y)
				x=x+y;
			
			else
				y=x+y;
			count++;
				
		}
	}

}
