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>
#include <bits/stdc++.h>
using namespace std;
unsigned long long mod_pow(unsigned long long num, unsigned long long pow,
                           unsigned long long mod) {
  unsigned long long test;
  unsigned long long n = num;
  for (test = 1; pow; pow >>= 1) {
    if (pow & 1) test = ((test % mod) * (n % mod)) % mod;
    n = ((n % mod) * (n % mod)) % mod;
  }
  return test;
}
long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int main() {
  int a, b;
  scanf("%d", &a);
  scanf("%d", &b);
  if (b > a)
    printf("%d ", -1);
  else {
    if (a == b) {
      double c = a;
      printf("%.12f", c);
    } else {
      double temp = a + b;
      double ans, zz;
      int st = 1, end = temp;
      zz = temp;
      ans = zz;
      while (st <= end) {
        int x = (st + end) / 2;
        zz = temp / (2 * x);
        if (zz >= b) {
          if (zz < ans) ans = zz;
          st = x + 1;
        } else {
          end = x - 1;
        }
      }
      printf("%.12f", ans);
    }
  }
}
