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;
 
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
 
#define nl '\n'
 
#define y1 aza_ibagaliev
 
#define f first
#define s second
	
#define ed end()
#define bg begin()
#define rbg rbegin()
 
#define rv reverse
#define mp make_pair
#define pb push_back
#define ppb pop_back
#define ppf pop_front
 
#define sz size()
#define all(x) x.bg, x.ed
 
#define pll pair < ll, ll >
#define pii pair < int, int >
 
#define sqr(x) ((x) * 1ll * (x))
#define sqrd(x) ((x) * 1.0 * (x))
 
#define speed ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
 
#define rnl cout << "----------------------------------------" << nl
 
#define bit __builtin_popcount
 
inline ll bip(ll x, ll n, ll mod){ll res=1;while(n){if(n & 1){res=(res*x)%mod;}x=(x*x)%mod;n>>=1;}return res;}
        	
const int ppr = 257;
const ll INF = 2e18;
const int inf = 2e9;
const int mod = 1e9 + 7;
const int N = 3e6 + 123;
const ld pi = 3.1741592653589793238462643;
 
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};

int n, a[25][25], dp[25][N];
vector<int>bt[25];

main() {             	
//  freopen(".in", "r", stdin);
//  freopen(".out", "w", stdout);

//	srand(time(NULL));

	scanf("%d", &n);

	for(int i = 1; i <= n; i++)
		for(int j = 0; j < n; j++)
			scanf("%d", &a[i][j]);
   
	for(int i = 0; i < (1<<n); i++) {
		int bi = bit(i);
		bt[bi].pb(i);
	}

	dp[0][0] = 1;
	for(int i = 1; i <= n; i++)
		for(int j = 0; j < n; j++) {
			if(!a[i][j]) continue;
			int bi = i - 1;
			for(auto x : bt[bi]) {
				if(x & (1<<j)) continue;
				dp[i][x|(1<<j)] = (dp[i][x|(1<<j)] + dp[i - 1][x]) % mod;
			}
		}

	cout << dp[n][(1<<n)-1];

    return 0;                                        
}