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)

#define _CRT_SECURE_NO_WARNINGS

#pragma GCC optimize("O3")

#include <iostream>
#include <algorithm>
#include <vector>
#include <cassert>
#include <cstring>
#include <numeric>
#include <set>
#include <queue>
#include <deque>
#include <array>
#include <string>
#include <cmath>
#include <random>
#include <ctime>
#include <map>
#include <functional>
#include <bitset>
#include <complex>
#include <iomanip>
#include <unordered_map>
#include <unordered_set>

#define sz(a) (int)((a).size())
#define all(a) (a).begin(), (a).end()
#define pb push_back

using namespace std;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
using ld = long double;

int bpc[1 << 21];
int dp[1 << 21];
int dp_new[1 << 21];

int main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	const int mod = 1e9 + 7;
	int n;
	cin >> n;
	for (int u = 1; u < (1 << n); u++) {
		bpc[u] = 1 + bpc[u & (u - 1)];
	}
	vector<vector<bool>> edge(n, vector<bool>(n));
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			int x;
			cin >> x;
			edge[i][j] = x;
		}
	}
	dp[0] = 1;
	for (int i = 0; i < n; i++) {
		memset(dp_new, 0, sizeof(dp_new));
		for (int mask = 0; mask < (1 << n); mask++) {
			if (bpc[mask] != i) continue;
			for (int j = 0; j < n; j++) {
				if (edge[i][j] && !(mask >> j & 1)) {
					int to = mask | (1 << j);
					dp_new[to] += dp[mask];
					if (dp_new[to] >= mod) {
						dp_new[to] -= mod;
					}
				}
			}
		}
		memcpy(dp, dp_new, sizeof(dp));
	}
	cout << dp[(1 << n) - 1];
	//cout << endl << 1.0 * clock() / CLOCKS_PER_SEC << endl;
}