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"
using namespace std;
using ll = long long;
using vll = std::vector<ll>;
using vvll = std::vector<vll>;
using vvvll = std::vector<vvll>;
using dd = double;
using vdd = std::vector<dd>;
using vvdd = std::vector<vdd>;
using vvvdd = std::vector<vvdd>;

constexpr ll INF = 1LL << 60;
constexpr dd EPS = 1e-11;

//cin,cout高速化のおまじない＋桁数指定
struct Fast{
	Fast(){
		cin.tie(0);
		ios::sync_with_stdio(false);
		cout << fixed << setprecision(std::numeric_limits<double>::max_digits10);
	}
} fast;

#define REPS(i, S, E) for (ll i = (S); i <= (E); i++)
#define REP(i, N) REPS(i, 0, N-1)
#define DEPS(i, S, E) for (ll i = (E); i >= (S); i--)
#define DEP(i, N) DEPS(i, 0, N-1)
#define EACH(e, v) for (auto&& e : v)

template<class T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true; }return false; }
template<class T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; }return false; }
template<class T> inline T MaxE(vector<T>&v,ll S,ll E){T m=v[S]; REPS(i,S,E)chmax(m,v[i]); return m;}//v[S]～v[E]の最大値
template<class T> inline T MinE(vector<T>&v,ll S,ll E){T m=v[S]; REPS(i,S,E)chmin(m,v[i]); return m;}//v[S]～v[E]の最小値
template<class T> inline T MaxE(vector<T> &v, ll N) { return MaxE(v, 0, N - 1); } //先頭N個中の最大値
template<class T> inline T MinE(vector<T> &v, ll N) { return MinE(v, 0, N - 1); }
template<class T> inline T MaxE(vector<T> &v) { return MaxE(v, (ll)v.size()); }
template<class T> inline T MinE(vector<T> &v) { return MinE(v, (ll)v.size()); }
template<class T> inline ll MaxI(vector<T>&v,ll S,ll E){ll m=S; REPS(i,S,E){if(v[i]>v[m])m=i;} return m;}
template<class T> inline ll MinI(vector<T>&v,ll S,ll E){ll m=S; REPS(i,S,E){if(v[i]<v[m])m=i;} return m;}
template<class T> inline ll MaxI(vector<T> &v, ll N) { return MaxI(v, 0, N - 1); }
template<class T> inline ll MinI(vector<T> &v, ll N) { return MinI(v, 0, N - 1); }
template<class T> inline ll MaxI(vector<T> &v) { return MaxI(v, (ll)v.size()); }
template<class T> inline ll MinI(vector<T> &v) { return MinI(v, (ll)v.size()); }
template<class T> inline T Sum(vector<T> &v, ll S, ll E){ T s=v[S]; REPS(i, S+1, E)s+=v[i]; return s; }
template<class T> inline T Sum(vector<T> &v, ll N) { return Sum(v, 0, N - 1); }
template<class T> inline T Sum(vector<T> &v) { return Sum(v, v.size()); }
template<class T> inline T POW(T a, ll n){ T r=1; for (; n>0; n>>=1, a*=a){ if (n&1)r*=a; } return r; }
inline ll POW(int a, ll n){ return POW((ll)a, n); }
inline ll MSB(ll a){for(ll o=63,x=-1;;){ll m=(o+x)/2; if(a<(1LL<<m))o=m; else x=m; if(o-x==1)return x;}}
inline ll CEIL(ll a, ll b){ return (a+b-1)/b; }

template<class T=ll> inline vector<T> cinv(ll N){ vector<T> v(N); REP(i, N)cin>>v[i]; return move(v);}
template<class T=ll, class S=ll> inline vector<pair<T, S>> cinv2(ll N){
	vector<pair<T, S>> v(N); REP(i,N){cin>>v[i].first>>v[i].second;} return move(v); }
template<class T=ll,class S=ll,class R=ll> inline vector<tuple<T, S, R>> cinv3(ll N){
	vector<tuple<T,S,R>> v(N); REP(i,N){cin>>get<0>(v[i])>>get<1>(v[i])>>get<2>(v[i]);} return move(v);}
template<class T=ll,class S=ll,class R=ll,class Q=ll> inline vector<tuple<T,S,R,Q>> cinv4(ll N){
	vector<tuple<T,S,R,Q>> v(N); REP(i,N){cin>>get<0>(v[i])>>get<1>(v[i])>>get<2>(v[i])>>get<3>(v[i]);} return move(v);}
template<class T> inline void coutv(vector<T> &v, char deli=' '){
	ll N=(ll)v.size(); REP(i,N){cout << v[i] << ((i==N-1)?'\n':deli);} }
template<class T> void bye(T a){cout << a << '\n'; exit(0);}


struct mll{
	static ll MOD;
	ll val;
	mll(ll v = 0): val(v % MOD){ if (val < 0) val += MOD; }
	mll operator - () const { return -val; }
	mll operator + (const mll &b) const { return val + b.val; }
	mll operator - (const mll &b) const { return val - b.val; }
	mll operator * (const mll &b) const { return val * b.val; }
	mll operator / (const mll &b) const { return mll(*this) /= b; }
	mll operator + (ll b) const { return *this + mll(b); }
	mll operator - (ll b) const { return *this - mll(b); }
	mll operator * (ll b) const { return *this * mll(b); }
	friend mll operator + (ll a, const mll &b) { return b + a; }
	friend mll operator - (ll a, const mll &b) { return -b + a; }
	friend mll operator * (ll a, const mll &b) { return b * a; }
	friend mll operator / (ll a, const mll &b) { return mll(a)/b; }
	mll &operator += (const mll &b) { val=(val+b.val)%MOD; return *this; }
	mll &operator -= (const mll &b) { val=(val+MOD-b.val)%MOD; return *this; }
	mll &operator *= (const mll &b) { val=(val*b.val)%MOD; return *this; }
	mll &operator /= (const mll &b) {
		ll c=b.val, d=MOD, u=1, v=0;
		while (d){
			ll t = c / d;
			c -= t * d; swap(c, d);
			u -= t * v; swap(u, v);
		}
		val = val * u % MOD;
		if (val < 0) val += MOD;
		return *this;
	}
	mll &operator += (ll b) { return *this += mll(b); }
	mll &operator -= (ll b) { return *this -= mll(b); }
	mll &operator *= (ll b) { return *this *= mll(b); }
	mll &operator /= (ll b) { return *this /= mll(b); }
	bool operator == (const mll &b) { return val == b.val; }
	bool operator != (const mll &b) { return val != b.val; }
	bool operator == (ll b) { return *this == mll(b); }
	bool operator != (ll b) { return *this != mll(b); }
	friend bool operator == (ll a, const mll &b) { return mll(a) == b.val; }
	friend bool operator != (ll a, const mll &b) { return mll(a) != b.val; }
	friend ostream &operator << (ostream &os, const mll &a) { return os << a.val; }
	friend istream &operator >> (istream &is, mll &a) { return is >> a.val; }
	static mll Combination(ll a, ll b){
		chmin(b, a-b);
		if (b<0) return mll(0);
		mll c = 1;
		REP(i, b) c *= a-i;
		REP(i, b) c /= i+1;
		return c;
	}
};
ll mll::MOD = 1000000007LL;// 998244353LL;//1000000007LL;(ll)(1e9 + 7);
using vmll = std::vector<mll>;
using vvmll = std::vector<vmll>;
using vvvmll = std::vector<vvmll>;
using vvvvmll = std::vector<vvvmll>;


void solve()
{
	ll N;
	cin >> N;
	vvll a(N);
	REP(i, N){ a[i] = cinv(N); }

	vector<map<ll, mll>> dp(N);

	REP(j, N){
		if (a[0][j] == 0) continue;
		dp[0][1LL<<j] = 1;
	}

	REPS(i, 1, N-1){
		EACH(e, dp[i-1]){
			ll bit = e.first;
			mll count = e.second;
			REP(j, N){
				if (a[i][j] == 0) continue;
				ll mask = (1LL<<j);
				if (bit & mask) continue;
				dp[i][bit | mask] += count;
			}
		}
	}
	mll ans = dp[N-1][(1LL<<N)-1];
	cout << ans << '\n';
}

int main(){
	solve();
	return 0;
}