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 CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
int gcd(int a, int b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
int phi(int n) {
  int res = 0;
  for (int i = 1; i <= n; i++) {
    if (gcd(i, n) == 1) res++;
  }
  return res;
}
int sigma(int n) {
  int res = 0;
  for (int i = 1; i <= n; i++)
    if (n % i == 0) res += i;
  return res;
}
int main() {
  int a, b;
  cin >> a >> b;
  if (b > a) {
    cout << -1;
    return 0;
  }
  if (a == b) {
    cout << a;
    return 0;
  }
  int k1 = (a - b) / (2 * b);
  int k2 = ((a + b) / (2 * b));
  long double x1 = (long double)(a - b) / (2 * k1);
  long double x2 = (long double)(a + b) / (2 * k2);
  if (x1 < x2)
    cout << setprecision(30) << x1;
  else
    cout << setprecision(30) << x2;
  return 0;
}