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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class C {
	
	static class Scanner{
		BufferedReader br=null;
		StringTokenizer tk=null;
		public Scanner(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		public String next() throws IOException{
			while(tk==null || !tk.hasMoreTokens())
				tk=new StringTokenizer(br.readLine());
			return tk.nextToken();
		}
		public int nextInt() throws NumberFormatException, IOException{
			return Integer.valueOf(next());
		}
		public long nextLong() throws NumberFormatException, IOException{
			return Long.valueOf(next());
		}
		public double nextDouble() throws NumberFormatException, IOException{
			return Double.valueOf(next());
		}
	}
	
	
	static double eps = 1e-9;
	
	
	static double sol(int a, int b) {
		// first case m = 1
		int Z = a - b;
		if (Z == 0)
			return a;
		int k = Z/ (2*b);
		return (Z/(2.0*k));
	}
	
	static double sol2(int a, int b) {
		// second case m = -1
		int Z = a + b;
		int k = Z / (2*b);
		return (Z/(2.0*k));
	}
	
	public static void main(String args[]) throws NumberFormatException, IOException {
		Scanner sc = new Scanner();
		int A = sc.nextInt();
		int B = sc.nextInt();
		double f = sol(A, B);
		double s = sol2(A, B);
		if (B > A)
			System.out.println("-1");
		else
			System.out.println(Math.min(f, s));
	}

}