Problem:
Score : 100 points

Problem Statement

There are N rabbits, numbered 1, 2, \ldots, N.

For each i, j (1 \leq i, j \leq N), the compatibility of Rabbit i and j is described by an integer a_{i, j}.
Here, a_{i, i} = 0 for each i (1 \leq i \leq N), and a_{i, j} = a_{j, i} for each i and j (1 \leq i, j \leq N).

Taro is dividing the N rabbits into some number of groups.
Here, each rabbit must belong to exactly one group.
After grouping, for each i and j (1 \leq i < j \leq N), Taro earns a_{i, j} points if Rabbit i and j belong to the same group.

Find Taro's maximum possible total score.

Constraints

All values in input are integers.

1 \leq N \leq 16

|a_{i, j}| \leq 10^9

a_{i, i} = 0

a_{i, j} = a_{j, i}

Input

Input is given from Standard Input in the following format:

N
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}

Output

Print Taro's maximum possible total score.

Sample Input 1

3
0 10 20
10 0 -100
20 -100 0

Sample Output 1

20

The rabbits should be divided as \{1, 3\}, \{2\}.

Sample Input 2

2
0 -10
-10 0

Sample Output 2

0

The rabbits should be divided as \{1\}, \{2\}.

Sample Input 3

4
0 1000000000 1000000000 1000000000
1000000000 0 1000000000 1000000000
1000000000 1000000000 0 -1
1000000000 1000000000 -1 0

Sample Output 3

4999999999

The rabbits should be divided as \{1, 2, 3, 4\}.
Note that the answer may not fit into a 32-bit integer type.

Sample Input 4

16
0 5 -4 -5 -8 -4 7 2 -4 0 7 0 2 -3 7 7
5 0 8 -9 3 5 2 -7 2 -7 0 -1 -4 1 -1 9
-4 8 0 -9 8 9 3 1 4 9 6 6 -6 1 8 9
-5 -9 -9 0 -7 6 4 -1 9 -3 -5 0 1 2 -4 1
-8 3 8 -7 0 -5 -9 9 1 -9 -6 -3 -8 3 4 3
-4 5 9 6 -5 0 -6 1 -2 2 0 -5 -2 3 1 2
7 2 3 4 -9 -6 0 -2 -2 -9 -3 9 -2 9 2 -5
2 -7 1 -1 9 1 -2 0 -6 0 -6 6 4 -1 -7 8
-4 2 4 9 1 -2 -2 -6 0 8 -6 -2 -4 8 7 7
0 -7 9 -3 -9 2 -9 0 8 0 0 1 -3 3 -6 -6
7 0 6 -5 -6 0 -3 -6 -6 0 0 5 7 -1 -5 3
0 -1 6 0 -3 -5 9 6 -2 1 5 0 -2 7 -8 0
2 -4 -6 1 -8 -2 -2 4 -4 -3 7 -2 0 -9 7 1
-3 1 1 2 3 3 9 -1 8 3 -1 7 -9 0 -6 -8
7 -1 8 -4 4 1 2 -7 7 -6 -5 -8 7 -6 0 -9
7 9 9 1 3 2 -5 8 7 -6 3 0 1 -8 -9 0

Sample Output 4

132

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define fi first
#define se second
#define mp make_pair
#define itrfor(itr,A) for(auto itr = A.begin(); itr !=A.end();itr++)
template <class T> using reversed_priority_queue = priority_queue<T, vector<T>, greater<T> >;
typedef long long llong;
char moji[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
char moji2[26]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
char moji3[10]={'0','1','2','3','4','5','6','7','8','9'};
#define Sort(a) sort(a.begin(),a.end());
#define Reverse(a) reverse(a.begin(),a.end());
#define print(a) cout << a << endl;
#define MOD llong(1e9+7)
#define MAX int(2 * 1e5 +5)
#define debug(x)  cout << #x << " = " << (x) << endl;
#define pi acos(-1.0)
#define int llong
#define INF llong(1e17)
template<class T> bool chmax(T &a,  T b) {if(a<b){a=b; return 1;} return 0;}
template<class T> bool chmin(T &a,  T b) {if(a>b){a=b; return 1;} return 0;}
void myprint(int* A,int A_num){
   REP(i,A_num) cout << A[i] << " ";
   cout << endl;
}

signed main(){
    int n;
    cin >> n;
    int joutaisuu = 1;
    REP(i,n) joutaisuu *= 2;
    int A[20][20];
    REP(i,n) REP(j,n) cin >> A[i][j];
    
    int dp[joutaisuu + 5];
    REP(i,joutaisuu) dp[i] = - INF;
    int hoge = 1;
    dp[0] = 0;
    int mask = 1;
    while(mask < joutaisuu){
        int mask2 = mask -1;
        mask2 = (mask2 & mask);
        while(mask2 > 0){
            int tmp = 0;
            tmp += dp[mask2];
            int mask3 = mask - mask2;
            tmp += dp[mask3];
            chmax(dp[mask],tmp);
            mask2 -= 1;
            mask2 = (mask2 & mask);
        }
        int tmp = 0;
        int id1 = 0;
        int hoge = 1;
        vector<int> ids;
        while(hoge < joutaisuu){
            if( (hoge & mask) ) ids.push_back(id1);
            hoge *= 2;
            id1 ++;
        }

        itrfor(itr1,ids){
            itrfor(itr2,ids){
                tmp += A[*itr1][*itr2];
            }
        }
        tmp /= 2;
        chmax(dp[mask],tmp);

        mask ++;

    }

    cout << dp[joutaisuu - 1] << endl;

}