Problem: Johnny needs to make a rectangular box for his physics class project. He has bought P cm of wire and S cm^2 of special paper. He would like to use all the wire (for the 12 edges) and paper (for the 6 sides) to make the box.
What is the largest volume of the box that Johnny can make?

Input
The first line contains t, the number of test cases (about 10). Then t test cases follow.

Each test case contains two integers P and S in a line (1 ≤ P ≤ 40000, 1 ≤ S ≤ 20000). You may assume that there always exists an optimal solution for the given input cases.


Output
For each test case, print a real number that is the largest volume of the box that Johnny can make, rounded to two decimal places.

Example

Input:
2
20 14
20 16

Output:
3.00
4.15

Output details
First case: the dimensions of the largest box may be 3, 1 and 1.
Second case: the dimensions of the largest box may be 7/3, 4/3 and 4/3.

Solution (in PYTHON):

Solution:
import math
def maxVol(p,s):
    d=(p*p*0.25)-(6*s)
    a1=((p/2)+math.sqrt(d))/6
    a2=((p/2)-math.sqrt(d))/6
 
    c1=(p/4)-(2*a1)
    c2=(p/4)-(2*a2)
 
    v=[(a1**2)*c1,(a2**2)*c2]
    return max(v)
    
for testcases in xrange(int(raw_input())) :
    perimeter, surface = map(int, raw_input().split() )
    print "%.2f"%maxVol(perimeter, surface)