CF div2 320 C

C. A Problem about Polyline

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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.

挺有意思的，在纸上推一遍就可以了，对于每个点(a,b)，显然如果b>a则无解,然后就是有解的情况，分两种情况，一种是点(a,b)在斜率为1的线上那么这个线段过（a-b,0）这个点，同理如果在斜率为-1的线上，那么这个线段过(a+b,0)这个点。然后开始各种推。

#include <cstdio>
#include <cstring>

using namespace std;

const int M = 10005;

int main()
{
    long long a,b;
    scanf("%lld%lld",&a,&b);
    if(a<b) puts("-1");
    else printf("%.12f
",(a+b)/(2.0*((a+b)/(2*b))));


    return 0