In this problem you will be given a half-circle. The half-circle’s radius is r. You can take any point A on the half-circle and draw 2 lines from the point to the two sides of the diameter(AB and AC). Let the sum of square of one line’s length and the other line’s length is s

Like in the figure s = AB² + AC. And BC = 2r.

Now given r you have to find the maximum value of s. That is you have to find point A such that AB² + AC is maximum.

Input

First line of the test case will be the number of test case T (1 <= T <= 1000). Then T lines follows. On each line you will find a integer number r (1 <= r <= 1000000); each representing the radius of the half-circle.

Output

For each input line, print a line containing "Case I: ", where I is the test case number and the maximum value of s. Print 2 digit after decimal (Errors should be less then .01).

Example

Sample Input:
1
1

Sample Output:
Case 1: 4.25

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	:): 2012-06-13 07:11:51 oh no i am getting TLE..used %Lf for answer and long long int for radius then also
	gabber: 2012-04-17 08:47:09 very easy just see the output formatting cost me three wa
	Darky: 2012-04-14 16:19:39 Somehow python gives wrong answer to this one.
	(Tjandra Satria Gunawan)(曾毅昆): 2012-03-24 18:20:53 i can solve this problem without using floating point or double... just integer with few math tricks :)
	ginnipkj: 2012-03-11 16:17:59 i totally agree with JVR and it is a bonus Question
	rohitjv: 2012-02-11 10:59:38 take care of the data types......i costed me many errors
	Jakub ©afin: 2012-01-02 19:32:58 Anyway, a testcase can be solved in constant time... the math behind it is nice and simple
	fandi a rahadian: 2011-11-30 11:40:32 why im getting TLE? lol. what will make me TLE, anybody please help
	MR. BEAN : 2011-10-19 13:46:15 simple differencial application!!!
	magicsquare: 2011-10-19 08:41:28 tutorial for sure!