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++ (GCC 9.2.1) (judge verdict: Accepted)

// #pragma GCC optimize ("O3")
// #pragma GCC target ("sse4")

#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 29;
const int NINF = -1*INF;
const long long LLINF = 1ll << 60;
const long long LLNINF = -1*LLINF;

template <class Ch, class Tr, class Container>
basic_ostream <Ch, Tr> & operator << (basic_ostream <Ch, Tr> & os, Container const& x) {
    os << "{ ";
    for(auto& y : x) os << y << " ";
    return os << "}";
}

template <class X, class Y>
ostream & operator << (ostream & os, pair <X, Y> const& p) {
	return os << "[ " << p.first << ", " << p.second << "]" ;
}

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
 
typedef pair<int, int> pi;
typedef pair<ll,ll> pll;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<double> vd;
typedef vector<ld> vld;
typedef vector<pi> vpi;
typedef vector<vector<int>> vvi;
typedef vector<vll> vvll;
typedef vector<vd> vvd;
typedef vector<vld> vvld;
 
#define mp make_pair
#define pb push_back
#define pf push_front
#define all(x) x.begin(), x.end()
#define revAll(x) x.rbegin(), x.rend()
#define gcd(a, b) __gcd(a, b)
#define popcount(x) __builtin_popcount(x)
#define isize(x) (int)(x.size())
#define rep(i,n) for(int i = 0; i < n; i++)
#define rep1(i, n) for(int i = 1; i < n+1; i++)
#define umap unordered_map

const int MOD = 1e9 + 7;
int madd(ll a, ll b){
	a += b;
	if (a >= MOD) a -= MOD;
	return a;
}

int msub(ll a, ll b){
	a -= b;
	if (a < 0) a += MOD;
	return a;
}


template <int MOD=1e9+7>
struct Modular {
  int value;
  static const int MOD_value = MOD;

  Modular(long long v = 0) { value = v % MOD; if (value < 0) value += MOD;}
  Modular(long long a, long long b) : value(0){ *this += a; *this /= b;}

  Modular& operator+=(Modular const& b) {value += b.value; if (value >= MOD) value -= MOD; return *this;}
  Modular& operator-=(Modular const& b) {value -= b.value; if (value < 0) value += MOD;return *this;}
  Modular& operator*=(Modular const& b) {value = (long long)value * b.value % MOD;return *this;}

  friend Modular mexp(Modular a, long long e) {
    Modular res = 1; while (e) { if (e&1) res *= a; a *= a; e >>= 1; }
    return res;
  }
  friend Modular inverse(Modular a) { return mexp(a, MOD - 2); }

  Modular& operator/=(Modular const& b) { return *this *= inverse(b); }
  friend Modular operator+(Modular a, Modular const b) { return a += b; }
  friend Modular operator-(Modular a, Modular const b) { return a -= b; }
  friend Modular operator-(Modular const a) { return 0 - a; }
  friend Modular operator*(Modular a, Modular const b) { return a *= b; }
  friend Modular operator/(Modular a, Modular const b) { return a /= b; }
  friend std::ostream& operator<<(std::ostream& os, Modular const& a) {return os << a.value;}
  friend bool operator==(Modular const& a, Modular const& b) {return a.value == b.value;}
  friend bool operator!=(Modular const& a, Modular const& b) {return a.value != b.value;}
};

int main(){
	ios_base::sync_with_stdio(false);cin.tie(0);
	int n; cin >> n;
	vvi A = vvi(n, vi(n, 0));
	rep(i, n){
		rep(j, n){
			int x; cin >> x;
			if (x) A[i][j] = 1;
		}
	}
	vector<Modular<MOD>> dp = vector<Modular<MOD>>(1 << n, 0);
	dp[0] = 1;
	for (int mask = 0; mask < (1 << n)-1; mask++){
		int next = popcount(mask);
		for (int cand = 0; cand < n; cand++){
			int bcand = 1 << cand;
			if (A[next][cand] and !(mask & bcand)){
				dp[mask ^ bcand] += dp[mask];
			}
		}
	}
	cout << dp[(1<<n)-1];
}