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>
#include <bits/stdc++.h>
using namespace std;
inline void rd(int &x) {
  x = 0;
  char c;
  while (c = getchar(), c < 48)
    ;
  do x = x * 10 + (c & 15);
  while (c = getchar(), c > 47);
}
void work(long long son, long long mom) {
  cout << (son / mom) << ".";
  son %= mom;
  for (int i = 1; i <= 12; ++i) {
    son *= 10;
    cout << son / mom;
    son %= mom;
  }
  cout << endl;
}
int main() {
  long long a, b;
  cin >> a >> b;
  if (a < b) {
    puts("-1");
    return 0;
  } else if (a == b) {
    cout << b;
    puts(".000000000000");
    return 0;
  }
  long long k = (a - b) / 2 / b;
  long long son = -1, mom = 1;
  if (k) {
    son = (a - b);
    mom = 1ll * 2 * k;
  }
  k = (a + b) / 2 / b;
  if (k >= 1) {
    if (son == -1 || 1ll * son * (2 * k) > 1ll * (a + b) * mom) {
      son = a + b;
      mom = 2 * k;
    }
  }
  if (~son)
    work(son, mom);
  else
    puts("-1");
  return 0;
}
