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>
const long long inf = std::numeric_limits<long long>::max();
using namespace std;
double x, y;
bool istrue(long long k) {
  double w = 2 * k;
  return (((x + y) / w >= y) || ((x - y) / w >= y));
}
long long bis(long long a, long long b) {
  if (a == b) return a;
  if (a == b - 1) {
    if (istrue(b))
      return b;
    else
      return a;
  }
  long long mid = (a + b) / 2;
  if (istrue(mid)) {
    return bis(mid, b);
  } else {
    return bis(a, mid);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> x >> y;
  if (bis(0, 1000000000) != 0) {
    double k = bis(0, 1000000000);
    if ((x + y) / (2 * k) >= y && (x - y) / (2 * k) < y) {
      cout << setprecision(32) << fixed << (x + y) / (2 * k);
    } else {
      cout << setprecision(32) << fixed << (x - y) / (2 * k);
    }
  } else {
    cout << -1;
  }
  return 0;
}