There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – .... 

We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.

Input

Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).

Output

Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output  - 1 as the answer.

Examples

Input

3 1


Output

1.000000000000


Input

1 3


Output

-1


Input

4 1


Output

1.250000000000

Note

You can see following graphs for sample 1 and sample 3. 

<image> <image>

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

/**
 * Created by sbabkin on 9/14/2015.
 */
public class SolverC {

    public static void main(String[] args) throws IOException {
        new SolverC().Run();
    }

    BufferedReader br;
    PrintWriter pw;
    StringTokenizer stok;

    private String nextToken() throws IOException {
        while (stok==null || !stok.hasMoreTokens()){
            stok = new StringTokenizer(br.readLine());
        }
        return stok.nextToken();
    }

    private int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    private long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    private void Run() throws IOException {
//        br=new BufferedReader(new FileReader("input.txt"));
//        pw=new PrintWriter("output.txt");
        br=new BufferedReader(new InputStreamReader(System.in));
        pw=new PrintWriter(new OutputStreamWriter(System.out));

        long a=nextInt();
        long b=nextInt();
        long x0=a+b;
        long count = x0/(b*2);
        if (count<=0){
            pw.println("-1");
        } else {
            pw.println((double)x0/(2*count));
        }
        pw.flush();
        pw.close();
    }

}
