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.Scanner;

public class Main {

	public static void main(String[] args) {
		
		BigInteger[] f = new BigInteger[200 + 2];
		f[0] = BigInteger.valueOf(1);
		f[1] = BigInteger.valueOf(1);
		for (int i = 2; i <= 200; ++i)
			f[i] = f[i - 1].add(f[i - 2]);
		Scanner in = new Scanner(System.in);
		BigInteger x, y, n;
		x = in.nextBigInteger();
		y = in.nextBigInteger();
		n = in.nextBigInteger();
		if (x.compareTo(n) >=0 || y.compareTo(n) >= 0)
			System.out.println(0);
		else if (x.compareTo(BigInteger.valueOf(0)) <= 0 && y.compareTo(BigInteger.valueOf(0)) <= 0)
			System.out.println(-1);
		else {
			BigInteger answer = BigInteger.valueOf(0);
			BigInteger max_ = x.max(y);
			BigInteger min_ = x.min(y);
			BigInteger a = min_.abs();
			if (min_ != a) {
				if (a.mod(max_).compareTo(BigInteger.valueOf(0)) == 0)
					min_ = max_;
				else
					min_ = a.divide(max_).add(BigInteger.valueOf(1)).multiply(max_).subtract(a);
				answer = a.divide(max_).add(BigInteger.valueOf(1));
			}
			//System.out.println(answer);
			for (int i = 1; i <= 200; ++i)
				if (max_.multiply(f[i]).add(min_.multiply(f[i - 1])).compareTo(n) >= 0) {
					answer = answer.add(BigInteger.valueOf(i));
					break;
				}
			System.out.println(answer);
		}
		in.close();
	}
}
