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;
template <typename T>
inline T sqr(T x) {
  T x_ = (x);
  return x_ * x_;
}
template <typename T>
inline T qbr(T x) {
  T x_ = (x);
  return x_ * x_ * x_;
}
template <typename T>
inline int sign(T x) {
  T x_ = (x);
  return ((x_ > T(0)) - (x_ < T(0)));
}
template <typename T>
inline T mod(T x, T m) {
  T x_ = (x);
  return (((x_) >= 0) ? ((x_) % (m)) : ((((x_) % (m)) + (m)) % (m)));
}
template <typename T>
inline T gcd(T a, T b) {
  while (b) {
    T t = a % b;
    b = a;
    a = t;
  }
  return a;
}
template <typename T>
inline T gcd_ex(T a, T b, T& x, T& y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  T x1, y1;
  T d = gcd_ex(b, a % b, x1, y1);
  x = y1;
  y = x1 - (a / b) * y1;
  return d;
}
template <typename T>
inline T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
template <typename T>
inline T binpow(T x, T deg) {
  T res = (T)1;
  for (; deg; x *= x, deg >>= 1)
    if (deg & 1) res *= x;
  return res;
}
template <typename T>
inline T modpow(T x, T deg, T m) {
  assert(deg >= (T)0);
  T res = (T)(1);
  for (; deg; x = (((x) % m) * ((x) % m)) % m, deg >>= 1)
    if (deg & 1) res = (((res) % m) * ((x) % m)) % m;
  return res;
}
template <typename T>
inline bool is_between(T POSITION, T LEFT, T RIGHT) {
  return (LEFT < POSITION) && (POSITION < RIGHT);
}
template <typename T>
inline bool is_inside(T POSITION, T LEFT, T RIGHT) {
  return (LEFT <= POSITION) && (POSITION <= RIGHT);
}
template <typename Collection, typename UnaryOperation>
void foreach (Collection& col, UnaryOperation op) {
  for_each(begin(col), end(col), op);
}
template <typename Collection, typename UnaryOperation>
Collection fmap(Collection& col, UnaryOperation op) {
  transform(begin(col), end(col), col.begin(), op);
  return col;
}
template <typename Collection, typename binop>
Collection zip(Collection& fc, Collection& sc, binop op) {
  transform(begin(fc), end(fc), sc.begin(), fc.begin(), op);
  return fc;
}
template <typename Collection, typename Condition>
bool exists(Collection& col, Condition con) {
  auto exist = find_if(begin(col), end(col), con);
  return exist != col.end();
}
template <typename Collection, typename Predicate>
Collection filterNot(Collection& col, Predicate predicate) {
  auto returnIterator = remove_if(begin(col), end(col), predicate);
  col.erase(returnIterator, end(col));
  return col;
}
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& P) {
  in >> P.first >> P.second;
  return in;
}
template <class T>
istream& operator>>(istream& in, vector<T>& Col) {
  for (auto& el : Col) in >> el;
  return in;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, vector<T>& Col) {
  for (auto& el : Col) os << el << " ";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, set<T>& Col) {
  for (auto& el : Col) os << el << " ";
  return os;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, map<T1, T2>& Col) {
  for (auto& el : Col) os << el << " ";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<vector<T>>& v) {
  for (auto& row : v) {
    for (auto& el : row) os << el << " ";
    os << "\n";
  }
  return os;
}
void solve() {
  int a, b;
  cin >> a >> b;
  if (b > a) {
    puts("-1");
    return;
  }
  int n = (a + b) / (2 * b);
  printf("%.9lf\n", (a + b + .0) / (2. * n));
}
int main() { solve(); }