Pie

My birthday is coming up and traditionally I’m serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. **This should be one piece of one pie, not several small pieces since that looks messy. **This piece can be one whole pie though.

My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.

What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.

Input

One line with a positive integer: the number of test cases. Then for each test case: —One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends. —One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies.

Output

For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10^(-3).

Sample Input

3 3 3 4 3 3 1 24 5 10 5 1 4 2 3 4 5 6 5 4 2

Sample Output

25.1327 3.1416 50.2655

這個題目的題意就是找出一個最大的實數(shù)V滿足：能從有的派中取出至少N+1個V，注意是取出而不能湊出。

如果考慮直接求解無疑是很麻煩的事，至少筆者自己想不出可以直接算出這個最大實數(shù)的方法，但是如果有一個數(shù)后確定這個數(shù)是不是滿足條件的就容易多了，只需要用V一一去除各蛋糕對應(yīng)的體積，把所得結(jié)果取整相加與N+1比較即可。 這個思路感覺沒什么好說的就直接上代碼把

代碼