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;

using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using ull = unsigned long long;
using vb = vector<bool>;
using vvb = vector<vb>;
using vd = vector<double>;
using vvd = vector<vd>;
using vvvd = vector<vvd>;
using vs = vector<string>;
using vvs = vector<vs>;
using vc = vector<char>;
using vvc = vector<vc>;
using ii = pair<int,int>;
using vii = vector<ii>;
using vvii = vector<vii>;
using dd = pair<double, double>;
using vdd = vector<dd>;
using vvdd = vector<vdd>;
using Mii = map<int,int>;
using vMii = vector<Mii>;

#define fi first
#define se second
#define INF INT_MAX
#define mod(a,m) (a%m+m)%m
#define PB push_back
#define MP make_pair

const int MOD = 1e9 + 7;

int n;
vvb v;
vvi memo;

int dp(int pos, int bits) {
	if(pos == n) return 1;
	int& ans = memo[pos][bits];
	if(ans != -1) return ans;
	ans = 0;
	for (int i = 0; i < n; ++i) {
		if(v[pos][i] and (bits & (1 << i))) {
			int new_bits = bits ^ (1 << i);
			ans = (ans + dp(pos+1, new_bits)) % MOD;
		}
	}
	return ans;
}

int main(){
//	cout.setf(ios::fixed);
//	cout.precision(4);
	while (cin >> n) {
		v = vvb(n, vb(n));
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < n; ++j) {
				int a;
				cin >> a;
				v[i][j] = a;
			}
		}
		memo = vvi(n, vi(1 << n, -1));
		cout << dp(0, (1 << n) - 1) << '\n';
	}
}