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)

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

#define ll long long
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define vll vector<ll>
#define vpii vector<pair<int,int>>
#define vpll vector<pair<ll,ll>>
#define fr(i,k,n) for (int i = k; i < n; ++i)
#define fri(i,k,n) for (int i = k; i >= n; --i)
#define INF (int)1e9
#define MOD 1000000007
#define pb push_back
#define mp make_pair
#define all(arr) arr.begin(),arr.end()

void boost(){
  ios_base::sync_with_stdio(false);
    cin.tie(NULL);
}

const int N=21,M=(1<<N);
ll dp[N][M];
int n,arr[N][N];

ll fun(int i,int mask){
  if(i!=__builtin_popcount(mask))
    return dp[i][mask]=0;
  if(dp[i][mask]!=-1)
    return dp[i][mask];
  ll ans=0;
  fr(j,0,n){
    if(arr[i][j]==1 && ((mask>>j)&1)==0){
      ans+=fun(i+1,mask^(1<<j));
      ans%=MOD;
    }
  }
  return dp[i][mask]=ans;
}

void solve(){
  cin>>n;
  fr(i,0,n){
    fr(j,0,n){
      cin>>arr[i][j];
    }
  }
  fr(i,0,N){
    fr(j,0,M){
      dp[i][j]=-1;
    }
  }
  fr(j,0,(1<<n))
  dp[n-1][j]=0;
  fr(j,0,n){
    if(arr[n-1][j]==1){
      dp[n-1][((1<<n)-1)^(1<<j)]=1;
    }
  }
  ll ans=0;
  cout<<fun(0,0);
  return;
}

int main()
{
  boost();
  
  int tc=1;
  //cin>>tc;
  while(tc--)
    solve();
  return 0;
}