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.io.*;
import java.util.*;

public class PerfectPair {
    public static void main(String[] args) throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(f.readLine());
        long x = Long.parseLong(st.nextToken());
        long y = Long.parseLong(st.nextToken());
        long m = Long.parseLong(st.nextToken());    
        if (x >= m || y >= m)
        {
            System.out.println(0);
            return;
        }
        if (x <= 0 && y <= 0)
        {
            System.out.println(-1);
            return;
        }
        if (x >= 0 && y >= 0)
            System.out.println(f(Math.min(x,y),Math.max(x,y),m));
        else
        {
            if (y < 0)
            {
                long temp = y;
                y = x;
                x = temp;
            }
            long k = -x/y;
            x += k*y;
            System.out.println(k+f(x,y,m));
        }
    }
    
    public static int f(long x, long y, long m)
    {
        if (y >= m)
            return 0;
        return 1 + f(Math.min(y, x+y),Math.max(y, x+y),m);
    }
}