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.
/**
 * Created by MalhotR1 on 05/18/2018.
 */

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class PerfectPairs {
    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            String[] in = br.readLine().trim().split(" ");
            long a = Long.parseLong(in[0]);
            long b = Long.parseLong(in[1]);
            long m = Long.parseLong(in[2]);
            if (a >= m || b >= m) System.out.println(0);
            else if (a <= 0 && b <= 0) System.out.println(-1);
            else {
                long count = 0;
                long[] arr = {Math.min(a,b), Math.max(a,b)};
                if (arr[0] < 0) {
                    long val = Math.abs(arr[0]);
                    count = val / arr[1];
                    if (val % arr[1] != 0) count++;
                    arr[0] += count*arr[1];

                    a = arr[0];
                    b = arr[1];
                }


                while (a < m && b < m) {
                    if (a < b) a += b;
                    else b += a;
                    count++;
                }
                System.out.println(count);

            }
        }
    }

}
