Problem:
Score : 100 points

Problem Statement

There are N men and N women, both numbered 1, 2, \ldots, N.

For each i, j (1 \leq i, j \leq N), the compatibility of Man i and Woman j is given as an integer a_{i, j}.
If a_{i, j} = 1, Man i and Woman j are compatible; if a_{i, j} = 0, they are not.

Taro is trying to make N pairs, each consisting of a man and a woman who are compatible.
Here, each man and each woman must belong to exactly one pair.

Find the number of ways in which Taro can make N pairs, modulo 10^9 + 7.

Constraints

All values in input are integers.

1 \leq N \leq 21

a_{i, j} is 0 or 1.

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 the number of ways in which Taro can make N pairs, modulo 10^9 + 7.

Sample Input 1

3
0 1 1
1 0 1
1 1 1

Sample Output 1

3

There are three ways to make pairs, as follows ((i, j) denotes a pair of Man i and Woman j):

(1, 2), (2, 1), (3, 3)

(1, 2), (2, 3), (3, 1)

(1, 3), (2, 1), (3, 2)

Sample Input 2

4
0 1 0 0
0 0 0 1
1 0 0 0
0 0 1 0

Sample Output 2

1

There is one way to make pairs, as follows:

(1, 2), (2, 4), (3, 1), (4, 3)

Sample Input 3

1
0

Sample Output 3

0

Sample Input 4

21
0 0 0 0 0 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 1
1 1 1 0 0 1 0 0 0 1 0 0 0 0 1 1 1 0 1 1 0
0 0 1 1 1 1 0 1 1 0 0 1 0 0 1 1 0 0 0 1 1
0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 0 0 0 1 1 0
1 1 0 0 1 0 1 0 0 1 1 1 1 0 0 0 0 0 0 0 0
0 1 1 0 1 1 1 0 1 1 1 0 0 0 1 1 1 1 0 0 1
0 1 0 0 0 1 0 1 0 0 0 1 1 1 0 0 1 1 0 1 0
0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 1 1 1 1 1 1
0 0 1 0 0 1 0 0 1 0 1 1 0 0 1 0 1 0 1 1 1
0 0 0 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 0 0 1
0 1 1 0 1 1 0 0 1 1 0 0 0 1 1 1 1 0 1 1 0
0 0 1 0 0 1 1 1 1 0 1 1 0 1 1 1 0 0 0 0 1
0 1 1 0 0 1 1 1 1 0 0 0 1 0 1 1 0 1 0 1 1
1 1 1 1 1 0 0 0 0 1 0 0 1 1 0 1 1 1 0 0 1
0 0 0 1 1 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1
1 0 1 1 0 1 0 1 0 0 1 0 0 1 1 0 1 0 1 1 0
0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 1 0 0 1
0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 0 0 1 1 0 1
0 0 0 0 1 1 1 0 1 0 1 1 1 0 1 1 0 0 1 1 0
1 1 0 1 1 0 0 1 1 0 1 1 0 1 1 1 1 1 0 1 0
1 0 0 1 1 0 1 1 1 1 1 0 1 0 1 1 0 0 0 0 0

Sample Output 4

102515160

Be sure to print the number modulo 10^9 + 7.

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

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

// clang-format off
#define  range(i, l, r) for (int i = (int)(l); i < (int)(r); ++(i))
#define rrange(i, l, r) for (int i = (int)(r) - 1; i >= (int)(l); --(i))

#define  whole(f, x, ...) ([&](decltype((x)) container) { return (f)(  begin(container),  end(container), ## __VA_ARGS__); })(x)
#define rwhole(f, x, ...) ([&](decltype((x)) container) { return (f)( rbegin(container), rend(container), ## __VA_ARGS__); })(x)

#define endl '\n'
#define debug(x) cerr << "(" << __LINE__ << ")" << #x << ": " << (x) << endl

using i32 = int;
using u32 = unsigned int;
using i64 = long long;
using u64 = unsigned long long;

// constexpr i32 mod   = 998244353;
constexpr i32 mod   = 1e9 + 7;
constexpr i32 inf   = 1001001001;
constexpr i64 infll = 1001001001001001001ll;

constexpr int dx[] = {0, -1, 1, 0, -1, 1, -1, 1}; 
constexpr int dy[] = {-1, 0, 0, 1, -1, -1, 1, 1};

struct IoSetup { IoSetup(int x = 15){ cin.tie(0); ios::sync_with_stdio(0); cout << fixed << setprecision(x); cerr << fixed << setprecision(x); } } iosetup;

template <typename T = i64> T input() { T x; cin >> x; return x; }

template <typename T> ostream &operator<<(ostream &os, vector<T> &v) { range(i, 0, v.size()) { os << v[i] << (i + 1 != v.size() ? " " : ""); } return os; } 
template <typename T> istream &operator>>(istream &is, vector<T> &v) { for (T &in : v) is >> in; return is; }

template <typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
template <typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { is >> p.first >> p.second; return is; }

template <typename T> vector<T> make_vector(size_t a, T b) { return vector<T>(a, b); }
template <typename... Ts> auto make_vector(size_t a, Ts... ts) { return vector<decltype(make_vector(ts...))>(a, make_vector(ts...)); }

template <typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
template <typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }
// clang-format on

void solver() {
    int N = input();
    auto a = make_vector(N, N, 0);
    cin >> a;
    
    auto dp = make_vector(1 << N, -1);

    function< int(int, int) > dfs = [&](int p, int bit) {
        if (bit + 1 == (1 << N)) return 1;
        if (p == N) return 0;
        
        int &res = dp[bit];
        if (~res) return res;
        res = 0;
        
        range(i, 0, N) {
            if ((1 << i) & bit || !a[p][i]) continue;
            (res += dfs(p + 1, bit | 1 << i)) %= mod;
        }
        
        return res;
    };
    
    cout << dfs(0, 0) << endl;
}

signed main(int argc, char *argv[]) {
    solver();
}