#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const ll mod = 1000000007;
#define si(x) scanf("%d",&x)
#define sil(x) scanf("%lld",&x)
#define pi(x) printf("%d\n",x)
#define pil(x) printf("%lld\n",x)
#define rep(i,n,m) for(int i = n; i< m; i++)
#define reps(i,n,m) for(int i = n; i >= m; i--)

int dp[17][(1<<17)],a[17][17],t,n,cas;

int Set(int N,int pos)
{
	//this fix a woman for a man
    return N = N | (1 << pos);
}

int Reset(int N,int pos)
{
    return N = N & ~(1 << pos);
}

bool check(int N,int pos)
{
	//this checks if this pos is empty or is not possessed by any man
    return (bool)(N & (1<<pos));
}
//bitmask dp
//bitmask dp is needed when all possible set is to be generated
int go(int men,int women)
{
	//base case is when all men are married
    if(men >= n) return 0;

    if(dp[men][women] != -1) return dp[men][women];

    int ret = 0;

   	for(int i = 0; i < n; i++){
        if(!check(women,i)){
			//if women "i" is single for this time for man "men" then we can arrange
			//a marriage and check whether it is in a set of of the maximum answer
            ret = max(ret,a[men][i] + go(men+1,Set(women,i)));
        }
    }

    return dp[men][women] = ret;

}
//it will always make monogamous family because we choose men one by one.
int main()
{
   	cin >> t;

    while(t--){
        cin >> n

        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                cin >> a[i][j];
            }
        }

        memset(dp,-1,sizeof(dp));

        int ans = go(0,0);

        printf("Case %d: %d\n",++cas,ans);
    }

    return 0;
}