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:
from sys import stdin

def findMaxVol(P, S):
	c = [(P/2.0+(P**2/4.0-6*S)**0.5)/6.0, (P/2.0-(P**2/4.0-6*S)**0.5)/6.0]
	v = [ce**3-P/4.0*(ce**2)+S/2.0*ce for ce in c]
	return max(v)

t=int(stdin.readline())
for _ in range(t):
	P, S = map(int, stdin.readline().split())
	print "%.2f"%findMaxVol(P, S)