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>
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int a, b;
  std::cin >> a >> b;
  if (a < b) {
    std::cout << "-1\n";
    return 0;
  }
  if (a == b) {
    std::cout << a << '\n';
    return 0;
  }
  long double ans = 1e12;
  {
    long long x = a + b;
    long long l = 1, r = (long long)4e9;
    while (l != r) {
      long long mid = (l + r + 1) / 2;
      if (x / mid / 2 / b >= 1) {
        l = mid;
      } else {
        r = mid - 1;
      }
    }
    ans = std::min(ans, (long double)x / l / 2.0);
  }
  {
    long long x = a - b;
    long long l = 0, r = (long long)4e9;
    while (l != r) {
      long long mid = (l + r + 1) / 2;
      if (x / mid / 2 / b >= 1) {
        l = mid;
      } else {
        r = mid - 1;
      }
    }
    if (l > 0) {
      ans = std::min(ans, (long double)x / l / 2.0);
    }
  }
  std::cout << std::fixed << std::setprecision(15) << (double)ans << '\n';
}