A circle is placed perfectly into a square. The term perfectly placed means that each side of the square is touched by the circle, but the circle doesn't have any overlapping part with the square. See the picture below.

Now you are given the radius of the circle. You have to find the area of the shaded region (blue part). Assume that pi = 2 * acos (0.0) (acos means cos inverse).

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case contains a floating point number r (0 < r ≤ 1000) denoting the radius of the circle. And you can assume that r contains at most four digits after the decimal point.

Output

For each case, print the case number and the shaded area rounded to two places after the decimal point.

Sample

Sample Input	Sample Output
Click to copy 3 20 30.091 87.0921	Click to copy Case 1: 343.36 Case 2: 777.26 Case 3: 6511.05

Notes

This problem doesn't have a special judge. So, precision problems could occur. Better to add a small value to your result to avoid this. For example, add 10^-9 to your result.

More about rounding errors.

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LOJ 1022 - Circle in Square

Input starts with an integer T (≤ 1000), denoting the number of test cases. Each case contains a floating point number r (0 < r ≤ 1000) denoting the radius of the circle. And you can assume that r contains at most four digits after the decimal point. For each case, print the case number and the shaded area rounded to two places after the decimal point.

Solution

We know that the area of a square = (length of any side)² and the area of a circle = π*(radius)². Here the length of any side of the square = 2*radius of the circle. We can easily calculate the area of the blue part

 area of the blue part = area of the square - area of the circle
                       = (2*r)² - π*r²

The above implementation is accepted.

Solution in C

#include <stdio.h>
#include <math.h>

int main()
{
    int t;
    double pi = 2.0 * acos(0.0);
    scanf("%d", &t);
    for (int i = 1; i <= t; i++)
    {
        double r;
        scanf("%lf", &r);
        double area = (2 * r * 2 * r) - (pi * r * r);
        printf("Case %d: %.2lf\n", i, area);
    }
    return 0;
}

Tutorial source