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 StartUp() {}
void StartUp(string in, string out) {
  freopen(in.c_str(), "r", stdin);
  freopen(out.c_str(), "w", stdout);
}
void StartUp(string str, string in, string out) {
  StartUp(str + "." + in, str + "." + out);
}
void StartUp(string str, bool def) {
  if (def) {
    StartUp(str, "in", "out");
  } else {
    StartUp(str, "gir", "cik");
  }
}
void StartUp(string str) { StartUp(str, false); }
template <typename T, typename U>
ostream& operator<<(ostream& out, pair<T, U>& v) {
  return out << "(" << v.first << "," << v.second << ")";
}
template <typename T>
void DBG(T x) {
  cerr << "x"
       << " =   " << x << endl;
}
template <typename T>
void DBG(T x, int n) {
  cerr << endl
       << "x"
       << " =" << endl;
  for (typeof(0) i = 0; i < n; i += 1) cerr << x[i] << endl;
}
template <typename T>
void DBG(T x, int n, bool tr) {
  if (tr)
    DBG(x, n);
  else {
    cerr << "x"
         << " =   ";
    for (typeof(0) i = 0; i < n; i += 1) cerr << x[i] << " ";
    cerr << endl;
  }
}
template <typename T>
void DBG(T x, int n, int m) {
  cerr << endl << "x" << endl;
  for (typeof(0) i = 0; i < n; i += 1) {
    for (typeof(0) j = 0; j < m; j += 1) {
      cerr << x[i][j] << " ";
    }
    cerr << endl;
  }
  cerr << endl;
}
int main() {
  StartUp();
  ios_base::sync_with_stdio(false);
  double a, b;
  scanf("%lf%lf", &a, &b);
  if (b > a) {
    cout << -1 << endl;
    return 0;
  }
  double c = (a - b), k = (int)c / (2 * (int)b);
  k *= 2;
  double ans = -1;
  if (k) ans = (ans == -1 ? (double)c / k : min(ans, (double)c / k));
  int t = a;
  t /= (int)b;
  if (t % 2 == 0) --t;
  k = t;
  for (typeof(0) i = 0; i < 1; i += 1) {
    c = a - b * k;
    if (c < 0) break;
    ans = (ans == -1 ? ((double)c / (k + 1)) + b
                     : min(ans, ((double)c / (k + 1)) + b));
    k -= 2;
  }
  printf("%.10lf\n", ans);
}