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;
using Int = long long;
template <typename T1, typename T2>
inline void chmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
struct FastIO {
  FastIO() {
    cin.tie(0);
    ios::sync_with_stdio(0);
  }
} fastio_beet;
template <typename T>
void drop(const T &x) {
  cout << x << endl;
  exit(0);
}
struct Precision {
  Precision() { cout << fixed << setprecision(12); }
} precision_beet;
signed main() {
  Int a, b;
  cin >> a >> b;
  if (a < b) drop(-1);
  using D = double;
  if (a == b) drop(D(a));
  D s = a + b, t = a - b;
  D ans = s / 2.0;
  Int p = (a + b) / (2 * b);
  chmin(ans, s / (2 * p));
  Int q = (a - b) / (2 * b);
  chmin(ans, t / (2 * q));
  cout << ans << endl;
  return 0;
}