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;
const int MOD = 1000000007;
long double PI = 4 * atan(1);
double x, b, a, ans = 1e9 + 7;
int l, r, mid;
double bin() {
  while (l < r) {
    mid = (l + r) / 2;
    if (x / (2 * mid) >= b) {
      l = mid + 1;
    } else {
      r = mid;
    }
  }
  return 2 * (l - 1);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> a >> b;
  l = 0, r = 1000000000, x = a + b;
  if (bin() > 0) ans = min(ans, (double)x / (double)bin());
  l = 0, r = 1000000000, x = a - b;
  if (bin() > 0) ans = min(ans, (double)x / (double)bin());
  if (ans == 1e9 + 7) {
    cout << fixed << setprecision(10) << -1;
  } else {
    cout << fixed << setprecision(10) << ans;
  }
  return 0;
}