encode01_17065
Arunava has learnt to find the sum of first N natural numbers using the expression N(N+1)/2.But now Kartik Sir has asked him to find the sum of floor(N/2) for first N natural numbers where floor(X) is the greatest integer less than or equal to a given real number X.Arunava has not memorized any formula for this. So can you help him. 


Input
1st line has an Integer T denoting the number of test cases
Each test Case Contains:
A single integer N on a new line denoting the number of natural number upto which it needs to be summed.



Output
For each test case output a single line containing required sum as the output



Example

Input:
5
1
2
3
4
5


Output:
0
1
2
4
6



CONSTRAINTS:
1 ≤ T  ≤ 100000
1 ≤ N ≤ 10000
0 ≤ S ≤ 214783647


Explanation of the fifth test case N=5,
The required sum is floor(1/2)+ floor(2/2)+ floor(3/2)+ floor(4/2)+ floor(5/2)=0+1+1+2+2=6
import sys

def main():
    answers = [0]*10000
    val = 0
    for i in xrange(1,5000):
        d = i*2
        val += d
        answers[d] = val-i
        answers[d+1] = val
    t = int(raw_input())
    inp = map(int, sys.stdin.read().split())
    out = ""
    for _ in inp:
        out+=str(answers[_])+"\n"
    print out[:-1]
main()