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:
#!/usr/bin/python
import math


def main():
    num_times = int(raw_input())
    for i in xrange(0, num_times):
        P, S = map(int, raw_input().split())
        l1 = ((P/2) + math.sqrt(math.pow(P/2, 2) - 6*S))/6
        l2 = ((P/2) - math.sqrt(math.pow(P/2, 2) - 6*S))/6
        h1 = P/4 - 2*l1
        h2 = P/4 - 2*l2
        v = max(h1 * math.pow(l1, 2), h2 * math.pow(l2, 2))
        print "%.2f" % v

if __name__ == "__main__":
    main()