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>
using namespace std;
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int a, b;
  cin >> a >> b;
  if (a < b) {
    cout << -1 << endl;
    return 0;
  }
  int p = (a + b - 1) / b;
  double x;
  if (a == b)
    x = b;
  else if (p % 2 == 0) {
    x = (double)(a + b) / (double)(p);
  } else {
    x = (double)(a - b) / (double)(p - 1);
    if (x < (double)b - 1e-9) x = (double)(a + b) / (double)(p - 1);
  }
  cout << fixed << setprecision(10) << x << endl;
}