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 read() { return; }
void print() { return; }
template <typename T1, typename... T2>
void read(T1 &x, T2 &...args) {
  ((cin >> x), read(args...));
}
template <typename T1, typename... T2>
void print(T1 x, T2... args) {
  ((cout << x << ' '), print(args...));
}
template <typename T>
void readArr(T &arr, int x, int y) {
  for (int i = (int)x; i < (int)y; ++i) cin >> arr[i];
}
template <typename T>
void printArr(T &arr, int x, int y) {
  for (int i = (int)x; i < (int)y; ++i) cout << arr[i] << ' ';
}
const int N = 1e6;
const int INF = 1e9;
const int MOD = 1e9 + 7;
int countBits(long long number) { return (int)log2(number) + 1; }
long long gcd(long long a, long long b) {
  return (b == 0) ? (a) : (gcd(b, a %= b));
}
long long add(long long x, long long y) { return (x + y) % MOD; }
long long sub(long long x, long long y) { return (x - y + MOD) % MOD; }
long long mul(long long x, long long y) { return (x * 1ll * y) % MOD; }
long long inv(long long p, long long q) {
  long long expo = MOD - 2;
  while (expo) {
    if (expo & 1) p = mul(p, q);
    q = mul(q, q), expo >>= 1;
  }
  return p;
}
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0) {
    long long tmp = power(x, y / 2);
    return mul(tmp, tmp);
  }
  return mul(x, power(x, y - 1));
}
void solveTestCase() {
  int a, b;
  read(a, b);
  if (a < b)
    print(-1);
  else
    cout << fixed << setprecision(12) << (a + b) / (2. * ((a + b) / (2 * b)));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t = 1;
  while (t--) solveTestCase();
  return 0;
}