#include <stdio.h>
#include <fstream>
using namespace std;

#define in "cast.in"
#define out "cast.out"
#define dim 13

int N, T, Tmin=-1;
int B[dim][1<<12-1], L[dim], C[dim];
bool Sel[dim];
int A[dim][dim];

int Maxim(int,int);
int Ok(int,int);
void Solve();

int main()
{
    freopen(in,"r",stdin);
    freopen(out,"w",stdout);
    
    scanf("%d", &T);
    for ( int k = 1; k <= T; k++ )
    {
        scanf("%d", &N);
        for ( int i = 1; i <= N; i++ )
            for ( int j = 1; j <= N; j++ )
                scanf("%d", &A[i][j]);
                
        Solve();
    }                
}    

int Maxim(int a, int b)
{
    if ( a > b ) return a;
    return b;
}    

int Minim(int a, int b)
{
    if ( a < b ) return a;
    return b; 
}

int Ok(int k1, int k2)
{
    for ( int i = 1; i <= N; i++ )
        if ( (k2>>(i-1))&1 ) 
             if ( !((k1>>(i-1))&1) ) return 0;
    
    return 1;  
}    

void Solve()
{
    int S = 1<<N;
    S -= 1;
    
    for ( int i = 1; i <= N; i++ )
        for ( int j = 1; j <= S; j++ )
            B[i][j] = 1000000;

    for ( int i = 1; i <= N; i++ )
        B[i][ 1<<(i-1) ] = 0;
   
   /* for ( int k1 = 1; k1 <= S; ++k1 )
    {
        for ( int i = 1; i <= N && 1<<(i-1) <= k1; ++i )
        {
            if ( (k1>>(i-1))&1 == 0 ) continue;
            for ( int k2 = 1; k2 < k1; ++k2 )
            {
                if ( !Ok(k1,k2) ) continue;
                for ( int j = 1; j <= N && 1<<(j-1) <= k2; ++j )
                {
                    if ( (k2>>(j-1))&1 == 0 || i == j ) continue;
                    
                    B[i][k1] = Minim( B[i][k1], A[i][j] + Maxim(B[i][k1-k2],B[j][k2]) ); 
                }             
            }
        }
    }*/
    
    int size=0;
    
    for ( int k1 = 1; k1 <= S; ++k1 )
    {
        for ( int i = 1; i <= N && 1<<(i-1) <= k1; ++i )
        {
            if ( (k1>>(i-1))&1 == 0 ) continue;
            
            if ( k1 == 1<<(i-1) ) continue;
            
            size = 0;
            
            for ( int j = 1; j <= N && 1<<(j-1) <= k1; ++j )
                if ( i != j && (k1>>(j-1))&1 ) size++, L[size] = j; 
            
            for ( int k2 = 1; k2 < (1<<size); ++k2 ) // sunt 2^size-1 posibilitati de a lua mult k1.
            {
                int t = 1, Q = 0;
                
                for ( ; t <= size && 1<<(t-1) <= k2; ++t )
                {
                      if ( (k2>>(t-1))&1 == 1 ) Q += 1<<(L[t]-1);
                }
                
              //  printf("%d ", Q);
                
                for ( int k = 1; k <= size && 1<<(k-1) <= k2; ++k )
                {
                    int j = L[k];
                    B[i][k1] = Minim( B[i][k1], A[i][j] + Maxim(B[i][k1-Q],B[j][Q]) );
                }   
            }
        }
    }
    
   /* printf("\n");
     for ( int i = 1; i <= N; i++, printf("\n") )
        for ( int j = 1; j <= (1 << N) - 1; j++ )
            printf("%d ", B[i][j]);*/
    
    printf("%d\n", B[1][S]);
}