#include <stdio.h>

#include <algorithm>

#include <cstring>

#include <cmath>

using namespace std;

#define met(a, b) memset(a, b, sizeof(a))

const double eps = 1e-;

const double PI = acos(-);

const int N = ;

struct point

{

    double x, y;

    point(double x=, double y=) : x(x), y(y){}

    friend point operator - (const point& p1, const point& p2)

    {

        return point(p1.x-p2.x, p1.y-p2.y);

    }

    friend double operator ^ (const point& p1, const point& p2)

    {

        return p1.x*p2.y - p1.y*p2.x;

    }

}p[N], res[N];

double Dist(point p1, point p2)

{

    double dx = p1.x - p2.x, dy = p1.y - p2.y;

    return sqrt(dx*dx + dy*dy);

}

bool cmp1(point p1, point p2)

{

    if(p1.y == p2.y)

        return p1.x < p2.x;

    return p1.y < p2.y;

}

bool cmp2(point p1, point p2)///极角排序；若极角相同，距离近的在前面；

{

    double k = (p1-p[])^(p2-p[]);

    if( k>eps || (fabs(k)<eps && Dist(p1, p[]) < Dist(p2, p[]) ))

        return ;

    return ;

}

int Graham(int n)

{

    res[] = p[];if(n == ) return ;

    res[] = p[];if(n == ) return ;

    int top = ;

    for(int i=; i<n; i++)

    {

        while(top && ((res[top]-res[top-])^(p[i]-res[top-])) <= ) top--;

        res[++top] = p[i];

    }

    return top+;

}

int main()

{

    int n, T, tCase = ;

    scanf("%d", &T);

    while(T--)

    {

        scanf("%d", &n);

        for(int i=; i<n; i++)

            scanf("%lf %lf", &p[i].x, &p[i].y);

        sort(p, p+n, cmp1);///p[0]为最下方靠左的点；

        sort(p+, p+n, cmp2);///以p[0]为基点，按叉积进行排序；

        int cnt = Graham(n);///求凸包的顶点个数cnt，保存在res中，下标从0开始；

        if(cnt < )

        {

            printf("Case %d: 0\n", tCase++);

            continue;

        }

        double ans = ;

        res[cnt] = res[], res[cnt+] = res[];

        for(int i=; i<=cnt; i++)

        {

            double a = Dist(res[i-], res[i+]);

            double b = Dist(res[i-], res[i]);

            double c = Dist(res[i], res[i+]);

            ans = min(ans, acos((b*b+c*c-a*a)/(*b*c)));

        }

        printf("Case %d: %.6f\n", tCase++, ans*/PI);

    }

    return ;

}