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>
const long long mod = (long long)1e9 + 7;
const int maxn = (int)1e5 + 1;
long long inf = 1LL << 60;
using namespace std;
inline void read(int &n) {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  n = x * f;
}
inline void read(long long &n) {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10ll * x + ch - '0';
    ch = getchar();
  }
  n = x * f;
}
pair<int, int> a[1000][1000];
int aa[1000][1000];
int ans[1000];
int main() {
  setlocale(LC_ALL, "Russian");
  ios_base::sync_with_stdio(0);
  int a, b;
  cin >> a >> b;
  if (a - b < 0) {
    puts("-1");
    return 0;
  }
  int f = a / b;
  if (f % 2 == 0) f--;
  cout << fixed << setprecision(10) << (a + b + 0.0) / (f + 1);
  return 0;
}