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 <class T>
inline void remax(T& A, T B) {
  if (A < B) A = B;
}
template <class T>
inline void remin(T& A, T B) {
  if (A > B) A = B;
}
string ToString(long long num) {
  string ret;
  bool neg = false;
  if (num < 0) num *= ~0, neg = true;
  do {
    ret += ((num % 10) + '0');
    num /= 10;
  } while (num);
  if (neg) ret += '-';
  reverse(ret.begin(), ret.end());
  return ret;
}
long long ToNumber(string s) {
  long long r = 0, p = 1;
  int e = (s[0] == '-');
  for (int i = s.size() - 1; i >= e; --i) r += (s[i] - '0') * p, p *= 10;
  if (e) r *= ~0;
  return r;
}
long long Gcd(long long a, long long b) {
  while (a %= b ^= a ^= b ^= a)
    ;
  return b;
}
long long Power(long long base, unsigned long long power) {
  long long ret = 1;
  while (power) {
    if (power & 1) ret *= base;
    power >>= 1;
    base *= base;
  }
  return ret;
}
long long PowerMod(long long base, unsigned long long power, long long mod) {
  if (!power) return 1 % mod;
  if (power & 1) return (base * PowerMod(base, power - 1, mod)) % mod;
  return PowerMod((base * base) % mod, power >> 1, mod);
}
int Log(long long num, long long base) {
  int ret = 0;
  while (num) {
    ++ret;
    num /= base;
  }
  return ret;
}
inline void run() {
  int a, b;
  scanf("%d%d", &a, &b);
  if (a < b) return (void)puts("-1");
  int k = (a + b) / b;
  if (k & 1) --k;
  printf("%.13f\n", (double)((long double)(a + b) / k));
}
int main() {
  FILE* input = stdin;
  FILE* output = stdout;
  while (!feof(input)) {
    run();
    break;
  }
  return 0;
}