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.*;
public class Main {
    static int[][] nums;
    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<B) {
             A^=B; B^=A; A^=B;
        }
        long ans = 0;
        if (B<0 && A>0 && A<M) {
            long ch = Math.min(-B/A,(-B+M+A-1)/A);
            ans += ch;
            B += ch*A;
        }
        while(A<M) {
            ans++;
            B = A+B;
            if (A<B) {
                A^=B; B^=A; A^=B;
            }
            if (ans > 3 && A<=0) {
                ans = -1;
                break;
            }
        }
        System.out.println(ans);
    }
}