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;
void prv(const vector<long long> &v) {
  long long n = v.size();
  for (long long i = 0; i < n; i++) cout << v[i] << " ";
  cout << "\n";
}
long long power(long long a, long long n, long long lelo) {
  if (n == 0) return 1;
  if (n == 1) return a;
  if (n == 2) return (a * a) % lelo;
  if (n % 2)
    return (a * power(a, n - 1, lelo)) % lelo;
  else
    return power(power(a, n / 2, lelo), 2, lelo) % lelo;
}
void swap(long long &a, long long &b) {
  long long tm = a;
  a = b;
  b = tm;
}
bool comp(const pair<long long, long long> &p1,
          const pair<long long, long long> &p2) {
  return ((p1.first + p1.second) < (p2.first + p2.second));
}
vector<string> split(const std::string &s, char delim) {
  vector<string> e;
  stringstream ss(s);
  string item;
  while (getline(ss, item, delim)) e.push_back(item);
  return e;
}
const long long N = 1e5 + 1;
void solve() {
  long long a, b;
  cin >> a >> b;
  if (a < b) {
    cout << -1;
    return;
  }
  if (a == b) {
    cout << a;
    return;
  }
  long long x = a - b, y = a + b;
  long long ans1, ans2;
  ans1 = x / (2 * b);
  ans2 = y / (2 * b);
  cout << fixed << setprecision(10);
  cout << min((1.0 * x / (2 * ans1)), (1.0 * y / (2 * ans2)));
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long(i) = (1); (i) < (t + 1); (i)++) {
    solve();
    cout << "\n";
  }
  return 0;
}