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 int max_m = 1010;
int n, m;
void input() {
  scanf("%d %d", &n, &m);
  return;
}
void solve() {
  if (n < m) {
    printf("-1");
    return;
  }
  if (n == m) {
    printf("%.12lf", (double)n);
    return;
  }
  long long f = (long long)(n - m), s = (long long)(n + m);
  long long div, div2;
  long long lo = 1, hi = 1000000000;
  while (lo < hi) {
    long long mid = (lo + hi + 1) / 2;
    if (f >= 1LL * 2 * mid * (long long)m) {
      lo = mid;
    } else
      hi = mid - 1;
  }
  div = lo * 2;
  lo = 1, hi = 1000000000;
  while (lo < hi) {
    long long mid = (lo + hi + 1) / 2;
    if (s >= 1LL * 2 * mid * (long long)m) {
      lo = mid;
    } else
      hi = mid - 1;
  }
  div2 = lo * 2;
  if (f < 1LL * div * m) {
    printf("%.12lf\n", (double)s / (double)div2);
    return;
  }
  printf("%.12lf\n",
         (double)min((double)f / (double)div, (double)s / (double)div2));
  return;
}
int main() {
  input();
  solve();
  return 0;
}