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)

#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
 
using namespace std;
// using namespace __gnu_pbds;
 
void err(istream_iterator<string> it) {cout << '\n';}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << " : ";
    err(++it, args...);
}
 
 
#define F first
#define S second
#define ll long long
#define pb push_back
#define db double
#define ld long double
#define ppb pop_back
#define pii pair<int, int>
#define pll pair<long long, long long>
#define piii pair<int, pii>
#define all(x) (x).begin(),(x).end()
#define __ ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define watch(x) cout << (#x) << " is " << (x) << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define TEST int T; cin >> T; while(T--)
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
//#define int long long
 
const int N = 1e5 + 5, inf = 2e9 + 7, mod = 1e9 + 7;
const ll INF = 2e18 + 7, MOD = 1e15 + 7;

int n, a[16][16];
ll pre[1 << 16], dp[1 << 16]; // dp[mask] = max ans if rabbits already taken

bool has(int n, int i) {
    return (n >> i) & 1;
}

void rec(int i, vector<int>& v, ll cur, int mask, int group) {
    if(i == v.size()) {
        dp[mask] = max(dp[mask], cur + pre[group]);
        return;
    }
    rec(i + 1, v, cur, mask ^ (1 << v[i]), group ^ (1 << v[i]));
    rec(i + 1, v, cur, mask, group);
}

signed main() { __
//    freopen("lcm.in", "r", stdin);
//    freopen("lcm.out", "w", stdout);
    cin >> n;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> a[i][j];
        }
    }
    
    for (int mask = 0; mask < (1 << n); mask++) {
        for (int i = 0; i < n; i++) {
            if (!has(mask, i)) continue;
            for (int j = i + 1; j < n; j++) {
                if (!has(mask, j)) continue;
                pre[mask] += a[i][j];
            }
        }
    }
//    for (int i = 0; i < (1 << n); i++) dp[i] = -INF;
    dp[0] = 0;
    for (int mask = 0; mask < (1 << n); mask++) {
        vector<int> v;
        for (int i = 0; i < n; i++) {
            if(!has(mask, i)) {
                v.pb(i);
            }
        }
        rec(0, v, dp[mask], mask, 0);
    }
    cout << dp[(1 << n) - 1];
    return 0;
}