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.*;
/* Codeforces 318C */

public class PerfectPair {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		long x = in.nextLong(), y = in.nextLong(), m = in.nextLong();
		
		System.out.println(minOperations(x, y, m));
	}
	
	static long minOperations(long x, long y, long m) {
		if (x >= m || y >= m) return 0;
		if (x <= 0 && y <= 0 && m > Math.max(x, y)) return -1;
		
		if (x >= 0 && y >= 0) { /* m > 0 */
			long count = 0;
			while (x < m && y < m) {
				long sum = x + y;
				y = Math.max(x, y);
				x = sum;
				count++;
			}
			return count;
		}
		
		/* xy < 0, m > 0 */
		if (y > 0) {
			long temp = x;
			x = y;
			y = temp;
		}
		
		long c = ceil(-y, x);
		y += c*x;
		return c + minOperations(x, y, m);
	}
	
	static long ceil(long x, long y) {
		if (x % y == 0) return x/y;
		return x/y + 1;
	}
}
