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;
const int inf = 0x3f3f3f3f;
const int maxn = 2e4 + 15;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
inline int read() {
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  int x = 0;
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  double x, y;
  cin >> x >> y;
  if (y > x) {
    cout << "-1" << endl;
    return 0;
  }
  double ans = inf;
  bool fg = false;
  int l = 0, r = inf;
  while (l <= r) {
    int mid = (l + r) >> 1;
    double res = (x - y) / (2 * mid);
    if (res >= y) {
      fg = true;
      ans = min(ans, res);
      l = mid + 1;
    } else
      r = mid - 1;
  }
  l = 0, r = inf;
  while (l <= r) {
    int mid = (l + r) >> 1;
    double res = (x + y) / (2 * mid);
    if (res >= y) {
      fg = true;
      ans = min(ans, res);
      l = mid + 1;
    } else
      r = mid - 1;
  }
  if (!fg)
    cout << "-1" << endl;
  else
    cout << fixed << setprecision(12) << ans << endl;
  return 0;
}
