Problem: 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>

Solution (in JAVA):

Solution:

import java.util.Scanner;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author hitarthk
 */
public class C_320c {
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        double a=in.nextDouble();
        double b=in.nextDouble();
        double min=Integer.MAX_VALUE;
        if(a>b){
            double f=(a-b)/2;
            //double p=Math.ceil(f);
            if(b==(a-b)/2){
                double ans=(a-b)/2;
                //System.out.println(ans);
                min=ans;
            }
            
            else{
//                int l=1;
//                int r=Integer.MAX_VALUE;
//                while(r-l>1){
//                    int mid=l+(r-l)/2;
//                    if((a-b)/(2*mid)==b){
//                        l=mid;
//                        break;
//                    }
//                    else if((a-b)/(2*mid)>b){
//                        l=mid;
//                    }
//                    else{
//                        r=mid;
//                    }
//                    
//                }
//                //System.out.println((a-b)/(2*l));
//                min=f/l;
                int p=(int)((a-b)/(2*b));
                min=f/p;
            }
                
        }
         if(a==b){
            System.out.println(a);
            System.exit(0);
        }
         {
            double f=(a+b)/2;
            if(b==(a+b)/2){
                //System.out.println((a+b)/2);
                if((a+b)/2<min)
                    min=(a+b)/2;
            }
            else if(b>f){
                System.out.println(-1);
                System.exit(0);
            }
            else{
//                int l=1;
//                int r=Integer.MAX_VALUE;
//                while(r-l>1){
//                    int mid=l+(r-l)/2;
//                    if((a+b)/(2*mid)==b){
//                        l=mid;
//                        break;
//                    }
//                    else if((a+b)/(2*mid)>b){
//                        l=mid;
//                    }
//                    else{
//                        r=mid;
//                    }
//                    
//                }
                //System.out.println((f/l));
                int p=(int)((a+b)/(2*b));
                if(f/p<min)
                    min=f/p;
            }
            
        }
         System.out.println(min);
    }
}