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.*;
import static java.lang.Math.*;
import java.io.*;

public class A {
	public static void p(Object...args) {System.out.println(Arrays.deepToString(args));}
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		long A = in.nextLong();
		long B = in.nextLong();
		long M = in.nextLong();
		if (A <= 0 && B <= 0) {
			if (A >= M || B >= M)
				System.out.println(0);
			else
				System.out.println(-1);
			return;
		}

		long ans = Long.MAX_VALUE;
		long moves = 0;
		while (A < M && B < M) {
			long a = min(A, B);
			long b = max(A, B);
			A = a;
			B = b;
			if (A > 0)
				ans = min(ans, moves + (M - B + (A-1)) / A);
			if (B > 0)
				ans = min(ans, moves + (M - A + (B-1)) / B);

			if (A < 0) {
				long times = (B - A) / B;
				A += times * B;
				moves += times;
			} else {
				A += B;
				moves++;
			}
		}
		ans = min(ans, moves);
		System.out.println(ans);
	}
}
