// Graph Gift
// Solution by Jacob Plachta

#define DEBUG 0

#include <algorithm>
#include <functional>
#include <numeric>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <complex>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <sstream>
using namespace std;

#define LL long long
#define LD long double
#define PR pair<int,int>

#define Fox(i,n) for (i=0; i<n; i++)
#define Fox1(i,n) for (i=1; i<=n; i++)
#define FoxI(i,a,b) for (i=a; i<=b; i++)
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
#define FoxR1(i,n) for (i=n; i>0; i--)
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
#define Min(a,b) a=min(a,b)
#define Max(a,b) a=max(a,b)
#define Sz(s) int((s).size())
#define All(s) (s).begin(),(s).end()
#define Fill(s,v) memset(s,v,sizeof(s))
#define pb push_back
#define mp make_pair
#define x first
#define y second

template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
template<typename T> T Sqr(T x) { return(x*x); }
string plural(string s) { return(Sz(s) && s[Sz(s)-1]=='x' ? s+"en" : s+"s"); }

const int INF = (int)1e9;
const LD EPS = 1e-12;
const LD PI = acos(-1.0);

#if DEBUG
#define GETCHAR getchar
#else
#define GETCHAR getchar_unlocked
#endif

bool Read(int &x)
{
	char c,r=0,n=0;
	x=0;
		for(;;)
		{
			c=GETCHAR();
				if ((c<0) && (!r))
					return(0);
				if ((c=='-') && (!r))
					n=1;
				else
				if ((c>='0') && (c<='9'))
					x=x*10+c-'0',r=1;
				else
				if (r)
					break;
		}
		if (n)
			x=-x;
	return(1);
}

#define LIM 30003

int NA,NB;
LL A[LIM],B[LIM];
LL SB[LIM];
LL mem3[LIM];

// cost for cross-pairing B values such that ny = n-y
LL Comp3(int ny)
{
		if (mem3[ny])
			return(mem3[ny]);
	int i;
	LL ans=0;
		Fox(i,(ny-1)/2)
			ans+=B[NB-i-2]*B[NB-ny+i];
	return(mem3[ny]=ans);
}

// min. cost for connecting last B values
LL Comp2(int n)
{
		if (!n)
			return(0);
		if ((n==1) && (NB>=2))
			return(B[NB-1]*B[NB-2]);
	int y;
	LL cur,ans=(LL)6e18;
	// pair first y values with the last value
		Fox1(y,n-1)
		{
				if ((n-y-1)%2)
					continue;
			cur=B[NB-1]*(SB[NB-n+y]-SB[NB-n]);
			// cross-pair the remaining n-y-1 values
			cur+=Comp3(n-y);
			Min(ans,cur);
		}
	return(ans);
}

// min. cost with last x A values paired with different B values
LL Comp1(int x)
{
	int i;
	LL ans=0;
		Fox(i,NA-x)
			ans-=B[0]*A[i];
		Fox(i,x)
			ans-=B[i+1]*A[NA-x+i];
	ans+=Comp2(NB-x-1);
	return(ans);
}

int main()
{
		if (DEBUG)
			freopen("in.txt","r",stdin);
	// vars
	int T,t;
	int N;
	int i,x;
	LL ans,v;
	// testcase loop
	Read(T);
		Fox1(t,T)
		{
			// input
			Read(N);
			NA=NB=0;
				while (N--)
				{
					Read(x);
						if (x>0)
							A[NA++]=x;
						else
							B[NB++]=-x;
				}
			// sort each list in non-increasing order
			sort(A,A+NA);
			sort(B,B+NB);
			reverse(A,A+NA);
			reverse(B,B+NB);
			// precompute prefix sums
				Fox(i,NB)
					SB[i+1]=SB[i]+B[i];
			// special case: all positive/negative?
			Fill(mem3,0);
				if ((!NA) || (!NB))
				{
						if (!NB)
						{
							NB=NA;
							memcpy(B,A,sizeof(A));
								Fox(i,NB)
									SB[i+1]=SB[i]+B[i];
						}
					ans=Comp2(NB);
				}
				else
				{
					// try all parameters
					ans=(LL)6e18;
						Fox(x,min(NA,NB))
							Min(ans,Comp1(x));
				}
			// output
			printf("Case #%d: %lld\n",t,ans);
		}
	return(0);
}