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 mp make_pair
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define ALL(a) begin(a), end(a)
#define SZ(a) ((int)(a).size())

#define REP(a, b) for (int a = 0; a < (b); a++)
#define FOR(a, b, c) for (int a = (b); a <= (c); a++)
#define REPD(a, b) for (int a = (b) - 1; a >= 0; a--)
#define FORD(a, b, c) for (int a = (b); a >= (c); a--)

#define MIN(a, b) (a) = min((a), (b))
#define MAX(a, b) (a) = max((a), (b))

#ifdef __DEBUG
#define debug if (true)
#else
#define debug if (false)
#endif

using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;

template <typename T> using V = vector<T>;
template <typename T> using VV = vector<V<T>>;
template <typename T> using VVV = vector<VV<T>>;

template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2> &p) {
  return is >> p.fi >> p.se;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  VV<int> mat(n, vi(n));
  REP(i, n) REP(j, n) cin >> mat[i][j];
  V<ll> sum(1 << n);
  REP(mask, (1 << n)) {
    REP(i, n) REP(j, i) if ((mask >> i & 1) && (mask >> j & 1)) sum[mask] += mat[i][j];
  }
  V<ll> dp(1 << n);
  REP(i, 1 << n) {
    dp[i] = sum[i];
    for (int j = (i - 1) & i; j > 0; j = (j - 1) & i) {
      MAX(dp[i], dp[j] + dp[i ^ j]);
    }
  }
  cout << dp[(1 << n) - 1] << endl;
}