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: Time Limit Exceeded)

#include<bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
#define int long long
#define x first
#define y second
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ve vector
#define forn(i,n) for(int i=0;i<(int)n;i++)
#define pb push_back
#define sz(a) (int)(a.size())
//using namespace __gnu_pbds;
using namespace std;

// // // functions: find_by_order  & order_of_key
// typedef tree<
// int,
// null_type,
// less<int>,
// rb_tree_tag,
// tree_order_statistics_node_update>
// ordered_set;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> ii;
typedef vector<ii> vii;
typedef vector<vi> vvi;

const int inf=(int)2e9;
const long long mod=1000000007; //1E9+7
int dx[]={1,-1,0,0},dy[]={0,0,1,-1};

ll powmod(ll a,ll p){
	ll ans=1;
	while(p){
		if(p&1) ans*=a;
		p/=2;a*=a; ans%=mod;a%=mod;
	}
	return ans;
}
ll mdinv(ll a){return powmod(a,mod-2); }


signed main(){
	IOS
	int n;cin>>n;
	vvi v(n,vi(n,0));
	forn(i,n) forn(j,n) cin>>v[i][j];
	vi DP(1<<21,0),DP_new(1<<21);
	DP[0]=1;
	for(int i=0;i<n;i++){
		for(int msk=0;msk<(1<<n);msk++){
			DP_new[msk]=0;
			for(int x=0;x<n;x++)
				if((msk>>x)&1 && v[i][x]){
					DP_new[msk]+=DP[msk^(1<<x)];
					if(DP_new[msk]>=mod)
						DP_new[msk]-=mod;
				}
		}
		swap(DP,DP_new);
	}
	cout<<DP[(1<<n)-1];
}