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>
#pragma comment(linker, "/STACK:256000000")
const int inf = 1e9 + 5;
const long long linf = 1e18;
const long double eps = 1e-10;
const long long mod = 1e9 + 7;
const int maxn = 2e5 + 10;
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  double ans;
  double a, b;
  cin >> a >> b;
  if (a < b) {
    cout << -1;
    exit(0);
  }
  if (a == b) {
    cout << a;
    exit(0);
  }
  double s1 = (a - b) / 2;
  double s2 = (a + b) / 2;
  int l = 1;
  int r = inf;
  while (l + 1 != r) {
    int m = (l + r) / 2;
    if (s1 / (double)m + eps < b) {
      r = m;
    } else {
      l = m;
    }
  }
  double ans1 = s1 / l;
  if (ans1 + eps < b) ans1 = inf;
  l = 1;
  r = inf;
  while (l + 1 != r) {
    int m = (l + r) / 2;
    if (s2 / (double)m + eps < b) {
      r = m;
    } else {
      l = m;
    }
  }
  double ans2 = s2 / l;
  cout.precision(10);
  cout << fixed << min(ans1, ans2);
  return 0;
}