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.InputStream;
import java.io.PrintStream;
import java.util.Scanner;


public class ProblemC {
	public static void main(String[] args) {
		ProblemC problem = new ProblemC();
		problem.solve(System.in, System.out);
	}

	public void solve(InputStream in, PrintStream out) {
		Scanner scan = new Scanner(in);
		long x = scan.nextLong();
		long y = scan.nextLong();
		long m = scan.nextLong();
		long a = Math.min(x, y);
		long b = Math.max(x, y);
		if ( b <= 0) {
			out.println(b >= m ? 0 : -1);
		} else {
			if ( b >= m) out.println(0);
			else {
				
				long steps = 0;
				if ( a < 0) {
					long absA = -a;
					steps += absA/b;
					a = -(absA % b);
				}
				while ( b < m ) {
					steps++;
					long z = a + b;
					if ( z > b ) {
						a = b;
						b = z;
					} else {
						a = z;
					}
				}
				out.println(steps);
			}
			
			
		}
	}
}
