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;
long long int x, y;
double mins, a, ans;
double binser(long long int awal, long long int akhir);
int main() {
  scanf("%I64d %I64d", &x, &y);
  if (x == y) {
    printf("%I64d\n", x);
    return 0;
  }
  if (x < y) {
    printf("-1\n");
    return 0;
  }
  mins = 1000000001;
  ans = binser(1, trunc(double(x / y)) + 1);
  if (ans == 1000000001) {
    printf("-1\n");
    return 0;
  }
  printf("%.*e\n", 9, ans);
}
double binser(long long int awal, long long int akhir) {
  long long int mid = (awal + akhir) / 2;
  if ((mid & 1 == 1) && (mid != 1)) {
    a = double(x - y) / (mid - 1);
  } else
    a = double(x + y) / mid;
  if (awal == akhir) return ((mins > a) && (a >= y) ? a : mins);
  if (a == double(y)) return a;
  if (a < double(y)) return binser(awal, mid - 1);
  if (mins > a) mins = a;
  if (a > double(y)) return binser(mid + 1, akhir);
}