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;
using ld = long double;
int is_eq(ld a, ld b) { return fabsl(a - b) < LDBL_EPSILON; }
ld f(ld sum, ld k) { return sum / (2 * k); }
int pred(ld r, ld y) { return is_eq(r, y) || r > y; }
int bin(ld sum, ld y) {
  int lo = 1, hi = 1e9;
  ld mx = f(sum, lo);
  if (!is_eq(mx, y) && mx < y) return -1;
  while (lo < hi) {
    int mid = (lo + hi) / 2;
    ld r = f(sum, mid);
    if (pred(r, y)) {
      if (lo == mid) {
        if (pred(f(sum, mid + 1), y))
          return mid + 1;
        else
          return mid;
      } else
        lo = mid;
    } else
      hi = mid - 1;
  }
  return lo;
}
ld get_ans(ld sum, ld y) {
  int k = bin(sum, y);
  if (k == -1) return INT_MAX;
  return f(sum, k);
}
int main() {
  int a, b;
  cin >> a >> b;
  if (b > a) {
    cout << -1 << endl;
    return 0;
  }
  if (a == b) {
    cout << fixed << setprecision(12) << ld(b) << endl;
    return 0;
  }
  ld ans1 = get_ans(a + b, b);
  ld ans2 = get_ans(a - b, b);
  ld ans = min(ans1, ans2);
  if (is_eq(ans, INT_MAX)) {
    cout << -1 << endl;
    return 0;
  }
  cout << fixed << setprecision(12) << ans << endl;
  return 0;
}
