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;
 
template <class T> ostream& operator << (ostream &os, const vector<T> &v) { for (T i : v) os << i << ' '; return os; }
template <class T> ostream& operator << (ostream &os, const set<T> &v) { for (T i : v) os << i << ' '; return os; }
template <class T, class S> ostream& operator << (ostream &os, const pair<T, S> &v) { os << v.first << ' ' << v.second; return os; }
template <class T, class S> ostream& operator << (ostream &os, const map<T, S> &v) { for (auto i : v) os << '(' << i.first << "=>" << i.second << ')' << ' '; return os; }
template <class T, class S> ostream& operator << (ostream &os, const unordered_map<T, S> &v) { for (auto i : v) os << '(' << i.first << "=>" << i.second << ')' << ' '; return os; }
 
#ifndef ONLINE_JUDGE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <class Arg1> void __f(const char* name, Arg1&& arg1) { cerr << name << " : " << arg1 << endl; }
    template <class Arg1, class... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args) {
        const char* sep = strchr(names + 1, ',');
        cerr.write(names, sep - names) << " : " << arg1 << "  ";
        __f(sep + 1, args...);
    }
#else
#define trace(...) 0
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("avx2,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _CRT_SECURE_NO_WARNINGS
#endif
 
#define fast_cin() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define MAX 1000000
#define ll long long
#define db double
#define str string
#define pb push_back
#define For(i,s,e) for (ll i=(s); i<(e); i++)
#define Forrev(i,s,e) for (ll i=(s); i>=(e); i--)
#define all(v) v.begin(),v.end()
 
#define vll vector<ll>
#define vs vector<string>
#define mapll map<ll,ll>
#define pll pair<ll,ll>
#define initialise(a, x) memset(a, x, sizeof(a))
#define maxheap priority_queue<ll>
#define minheap priority_queue<ll,vector<ll> ,greater<ll>>
 
#define ff first
#define ss second
#define endl "\n"
#define mp make_pair
const ll mod=1e9 + 7;
 
ll takemod(ll a) {
    return ((a%mod)+mod)%mod;
}
 
ll gcd(ll a, ll b) { 
    if (b == 0) 
        return a; 
    return gcd(b, a % b);        
}
 
ll fast_exp(ll base, ll expo) {
    ll res=1;
    while(expo>0) {
        if(expo&1) res=(res*base)%mod;
        base=(base*base)%mod;
        expo>>=1;}
    return res;
}
 
ll modinv(ll a) {
    return takemod(fast_exp(takemod(a), mod-2));
}
 
const int N = (1<<21);

int dp[N], arr[21][21];
int n;

int dpfunc(int val) {
	if (__builtin_popcountll(val) == n)
		return 1;
	else if (dp[val] != -1)
		return dp[val];
	
	int ans = 0, index = __builtin_popcountll(val);
	for (int i = 0; i < n; i++) {
		if ((val & (1<<i)) == 0 and arr[index][i] == 1)
			ans = (ans + dpfunc(val + (1<<i)))%mod; 
	}
	dp[val] = ans;
	return dp[val];
}

void solve () {
	
	cin >> n;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++)
			cin >> arr[i][j];
	}
  
	memset(dp, -1, sizeof(dp));
	cout << dpfunc(0) << endl;
}


 
signed main() 
{
    fast_cin();
    
    int t=1;
    // cin>>t;
    while(t--)
    {
        solve();
    }
    
    
    return 0;    
}