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;
import static java.lang.Math.*;

public class A {
	public static int minimum(long a, long b, long m) {
		BigInteger aa = BigInteger.valueOf(a);
		BigInteger bb = BigInteger.valueOf(b);
		BigInteger mm = BigInteger.valueOf(m);
		for (int i = 0; i < 512; i++) {
			if (aa.compareTo(mm) >= 0 || bb.compareTo(mm) >= 0)
				return i;
			if (i % 2 == 0) {
				if (bb.add(aa).compareTo(mm) >= 0)
					return i + 1;
				aa = aa.add(bb);
			} else {
				if (aa.add(bb).compareTo(mm) >= 0)
					return i + 1;
				bb = bb.add(aa);
			}
		}
		return -1;
	}

	public static long solve(long x, long y, long m) {
		if (x >= m || y >= m)
			return 0;
		if (x >= 0 && y >= 0)
			return min(minimum(x, y, m), minimum(y, x, m));
		if (x <= 0 && y <= 0)
			return -1;
		long max = max(x, y);
		long min = min(x, y);
		long turns = -min / max + (((-min) % max) == 0 ? 0 : 1);
		return turns + solve(max, turns * max + min, m);
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		long x = in.nextLong();
		long y = in.nextLong();
		long m = in.nextLong();
		System.out.println(solve(x, y, m));
	}
}
