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.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author foo.coder
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskC {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
        int a = in.nextInt();
        int b = in.nextInt();

        if ( b > a ) {
            out.println(-1);
        } else if (a % b == 0 && (a / b) % 2 == 1) {
            out.printf("%.12f", (double)b);
        } else {
            double n1 = (a - b) / (2*b);
            double x1 = (a - b) / (2.0*n1);

            double n2 = ((a + b) - 2*b) / (2 * b);
            double x2 = (a + b) / (2.0*(n2 + 1));

            out.printf("%.12f", Math.min(x1, x2) );
        }
    }
}

