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.*;
import java.io.*;

public class C {
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        StringTokenizer st = new StringTokenizer(f.readLine());
        long a = Integer.parseInt(st.nextToken());
        long b = Integer.parseInt(st.nextToken());
        long p1 = a-b;
        long p2 = a+b;
        long lo = 1;
        long hi = 1000000007L;
        double ans = Double.MAX_VALUE;
        if(p1 < 0){
            out.println(-1);
            out.close();
            System.exit(0);
        }
        if(p1 == 0){
            out.println(b);
            out.close();
            System.exit(0);
        }
        if(p1 > 0) {
            while (lo < hi-1) {
                long mid = (hi + lo) / 2;
                double val = Math.round((((double)p1 / (2.0*(double)mid))*1000000000.0))/1000000000.0;
                if(val < b){
                    hi = mid;
                }else{
                    lo = mid;
                }

            }
        }
        double nval = Math.round((((double)p1 / (2.0*(double)lo))*1000000000.0))/1000000000.0;
        if(nval >= b) ans = Math.min(ans, nval);

        lo = 1;
        hi = 1000000007L;
        if(p2 > 0) {
            while (lo < hi-1) {
                long mid = (hi + lo) / 2;
                double val = Math.round((((double)p2 / (2.0*(double)mid))*1000000000.0))/1000000000.0;
                if(val < b){
                    hi = mid;
                }else{
                    lo = mid;
                }

            }
        }
        nval = Math.round((((double)p2 / (2.0*(double)lo))*1000000000.0))/1000000000.0;
        if(nval >= b) ans = Math.min(ans, nval);

        if(ans == Double.MAX_VALUE){
            out.println(-1);
        }else {
            out.println(ans);
        }
        out.close();




    }
}
