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)

#ifdef _MSC_VER
#include "stdc++.h"
#endif
#ifdef __GNUC__
#include<bits/stdc++.h>
#endif
using namespace std;

using ll = long long;

//constant
const double EPS = 1e-10;

#define REP(i, n) for (ll i = 0; i < n; i++)
#define FOR(i, s, e) for (ll i = s; i < e; i++)
#define RREP(i, n) for (ll i = n; i >= 0; i--)
#define ALL(x) (x).begin(),(x).end()
#define RALL(x) (x).rbegin(), (x).rend()

//debug
#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define DEBUG(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl;

// vector
//template<typename T> vector<vector<T>> vv(ll H, ll W, T v) { return vector<vector<T>> vec(H, vector<T>(W, v)); }

// view vector
template<typename T> void view(T e) { cerr << e << endl; }
template<typename T> void view(const vector<T>& v) { for (const auto& e : v) { cerr << e << " "; } cerr << endl; }
template<typename T> void view(const vector<vector<T> >& vv) { for (const auto& v : vv) { view(v); } }

// DP
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }


const ll MOD = 1000000007;
const ll inf = 1LL << 60;

struct Mint {
    ll v;
    ll _mod;
    Mint() : v(0) {}
    Mint(signed v, ll _mod = MOD) : v(v), _mod(_mod) {}
    Mint(ll t, ll _mod = MOD) : _mod(_mod) {
        v = t % _mod;
        if (v < 0)
            v += _mod;
    }

    Mint pow(ll k) {
        Mint res(1), tmp(v);
        while (k) {
            if (k & 1)
                res *= tmp;
            tmp *= tmp;
            k >>= 1;
        }
        return res;
    }
    static Mint add_identity() { return Mint(0); }
    static Mint mul_identity() { return Mint(1); }
    Mint inv() { return pow(_mod - 2); }

    Mint& operator+=(Mint a) {
        v += a.v;
        if (v >= _mod)
            v -= _mod;
        return *this;
    }
    Mint& operator-=(Mint a) {
        v += _mod - a.v;
        if (v >= _mod)
            v -= _mod;
        return *this;
    }
    Mint& operator*=(Mint a) {
        v = v * a.v % _mod;
        return *this;
    }
    Mint& operator/=(Mint a) { return (*this) *= a.inv(); }

    Mint operator+(Mint a) const { return Mint(v) += a; };
    Mint operator-(Mint a) const { return Mint(v) -= a; };
    Mint operator*(Mint a) const { return Mint(v) *= a; };
    Mint operator/(Mint a) const { return Mint(v) /= a; };

    Mint operator-() const { return v ? Mint(_mod - v) : Mint(v); }

    bool operator==(const Mint a) const { return v == a.v; }
    bool operator!=(const Mint a) const { return v != a.v; }
    bool operator<(const Mint a) const { return v < a.v; }
};
ostream& operator<<(ostream& os, Mint m) { return os << m.v; }

Mint nCk(int n, int k) {
    Mint ret = 1;
    for (int i = n; i > n - k; --i) {
        ret *= i;
    }
    for (int i = 1; i <= k; ++i) {
        ret /= i;
    }
    return ret;
}
// ***************************************

int main() {

    int N;
    cin >> N;

    vector<Mint> dp((1 << N) + 1, 0);
    vector<vector<bool>> A(N, vector<bool>(N, false));
    int a;
    REP(i, N) REP(j, N) {
        cin >> a;
        A[i][j] = a;
    }

    dp[0] = 1;

    REP(S, 1 << N) {
        int cnt = 0;
        REP(j, N) {
            if (S & (1 << j)) cnt++;
        }

        REP(j, N) {
            if ((S & (1 << j) && A.at(cnt - 1)[j])) {
                dp[S] += dp[S ^ (1 << j)];
            }
        }
    }

    cout << dp[(1 << N) - 1].v << endl;

#ifdef _MSC_VER
    system("PAUSE");
#endif
    return 0;
}