{"trustable":false,"sections":[{"title":"","value":{"format":"MD","content":"\u003cp\u003eImagine you have got a circle in your final exam! Sad? Okay, then you have got two circles after 1. Now you are happy? So take one circle from your result sheet and placed the circle perfectly into a square. The term perfectly placed means that the circle is touching the each side of the square, but the circle doesn\u0027t have any overlapping part with the square. See the picture below.\u003c/p\u003e\n\n\u003cimg src\u003d\"https://i.postimg.cc/yNRcY705/Screenshot-2021-11-22-203131.png\" alt\u003d\"Circle in a Square\" width\u003d\"200\" height\u003d\"200\"\u003e\u003cbr\u003e\n\n\u003cp\u003eNow you need to find the area of the shaded region (blue part) after giving the radius. Assume that pi \u003d 2 * acos (0.0) (acos means cos inverse).\u003c/p\u003e\n\n\u003ch4\u003eInput\u003c/h4\u003e\n\n\u003cp\u003eThe number of test cases is denoted by T(≤ 1000) which is an integer number and the input starts with it. A floating point number r is contained by each case where r is radius and 0 \u003c r ≤ 1000. And you can assume that r contains at most four digits after the decimal point.\u003c/p\u003e\n\n\u003ch4\u003eOutput\u003c/h4\u003e\n\n\u003cp\u003eFor all the cases, print the case number. After printing the case number you should print the shaded area rounded to two places after the decimal point.\u003c/p\u003e\n\n\u003ch4\u003eSample Input\u003c/h4\u003e\n\n3\u003cbr\u003e\n20\u003cbr\u003e\n30.091\u003cbr\u003e\n87.0921\n\u003ch4\u003eSample Output\u003c/h4\u003e\n\nCase 1: 343.36\u003cbr\u003e\nCase 2: 777.26\u003cbr\u003e\nCase 3: 6511.05"}},{"title":"","value":{"format":"MD","content":"\u003ch4\u003eNote\u003c/h4\u003e\nYou can submit your solution using C/C++/Java/JavaScript"}}]}