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)

/*
___mennat khoday ra azz o va jall ke taatash mojebe ghorbat ast o be shokr andarash mazide nemat.
___har nafasi ke foro miravad momedde hayatast o chon bar miayad mofarrahe zat.
___pas dar har nafasi 2 nemat mojod ast o bar har nematy shokry vajeb.

___az dast o zabane ke barayad ?
___kaz ohdeye shokrash be dar ayad !


___man be to dami gharar natvanam kard
___va ehsane to ra shomar natvanam kard
####################
___gar bar tane man zaban shavad har moyee
___yek shokre to az hezar natvanam kard

*/

#include <bits/stdc++.h>

#define pb push_back
#define SALI main
#define left(x) ((x)*2)
#define right(x) (((x)*2)+1)
#define fuck(x) cout << #x << " : " << x << endl;
#define findLowVec(v,x) (lower_bound(v.begin(), v.end(), x) - v.begin())
#define findUpVec(v,x) (upper_bound(v.begin(), v.end(), x) - v.begin())
#define findLowArr(a,n,x) (lower_bound(a, a+n, x) - a)
#define findUpArr(a,n,x) (upper_bound(a, a+n, x) - a)
#define int long long

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef long double ld;

template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p)
{return out << "(" << p.first << ", " << p.second << ")";}

template<class A> ostream& operator <<(ostream& out, const vector<A> &v)
{out << "[";for(int i = 0; i < v.size(); i++) {if(i) out << ", ";out << v[i];}return out << "]";}

template<class A,class B> ostream& operator <<(ostream& out, const set<A,B> &s)
{out << "[";for(auto i = s.begin(); i != s.end(); i++) {out << ", ";out << *i;}return out << "]";}

template<class A> ostream& operator <<(ostream& out, const set<A> &s)
{out << "[";for(auto i = s.begin(); i != s.end(); i++) {out << ", ";out << *i;}return out << "]";}

const int N = 22;
const int N2 = 1<<22;
const int mod = 1e9 + 7;
int n;
int dp[N][N2];
int a[N][N];



int solve(int id, int bit)
{
    if(id == -1) return 1;
    if(dp[id][bit] != -1) return dp[id][bit];
    int ans = 0;
    for(int i = 0; i < n; i++){
        if(a[id][i] == 1 && !(bit & (1 << i))){
            ans += solve(id-1, bit|(1<<i) );
            ans %= mod;
        }
    }
    return dp[id][bit] = ans;

}

int32_t SALI()
{

    memset(dp,-1,sizeof(dp));
    cin >> n;
    for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) cin >> a[i][j];

    cout << solve(n-1,0);
}




/**< WRITEN BY ALI ADELKHAH */