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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	ash_demon8: 2018-04-05 18:13:27 AC in one go...... with 4 lines of code:)
	abhinav__: 2018-04-02 17:31:08 Use double as the data type and make sure to type cast your solution when printing
	prateek_imkp1: 2018-03-30 07:16:32 Use long long int _/\_
	ayusofayush: 2018-02-23 14:18:51 simple differentiation and use setprecision() ....only maths
	Divyam Shah: 2018-02-09 17:24:19 @Nani the value of roots is not imaginary, for AC=0.5 we get "AC^2 - AC = -0.25".
	vkash: 2018-01-03 10:55:20 4 lines of code...easy 1 :)
	nani: 2017-12-27 19:37:56 There seems to be some loophole in the question, pythagorean theorem AB^2 + AC^2 = BC^2. For sample input AB^2 + AC^2 = 4, but how will AB^2 + AC = 4.25 this tends to be AC^2 - AC = -0.25 which in return leads to imaginary values of AC. Please clarify me if I am missing anything?
	techventer_936: 2017-06-30 23:06:05 Simple AC in one go :) look for input data type and precision.
	Anand Kumar: 2017-06-25 10:57:14 took time.. but finally got it...
	shalini6: 2017-06-08 14:06:41 Use s = AB^2 + AC and pythagoras theorem AB^2 + AC^2 = (2r)^2(traingle in a semi circle is right angled) Then differentiate. Simple maxima