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;
const double PI = 3.1415926535897932384626433;
const long long mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const double eps = 1e-12;
template <class T>
T gcd(T a, T b) {
  if (!b) return a;
  return gcd(b, a % b);
}
const int maxn = 1e3 + 100;
long long a, b;
double ans;
int solve() {
  if (a == b) {
    ans = b;
    return 1;
  }
  int flg = 0;
  ans = 1e9 + 100;
  if (a >= b * 3) {
    flg = 1;
    long long l = 1, r = 1e9 + 10;
    while (l + 1 < r) {
      if (a >= (2 * ((l + r) / 2) + 1) * b)
        l = ((l + r) / 2);
      else
        r = ((l + r) / 2);
    }
    ans = min(ans, (a - b) * 1.0 / 2.0 / l);
  }
  if (a >= b) {
    flg = 1;
    long long l = 1, r = 1e9 + 10;
    while (l + 1 < r) {
      if (a >= (2 * ((l + r) / 2) - 1) * b)
        l = ((l + r) / 2);
      else
        r = ((l + r) / 2);
    }
    ans = min(ans, (a + b) * 1.0 / 2.0 / l);
  }
  return flg;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  while (cin >> a >> b) {
    if (solve())
      printf("%.10f\n", ans);
    else
      printf("-1\n");
  }
  return 0;
}