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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define lli long long int
#define vec2 vector<vector<lli>>
#define fio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define FRE freopen("explicit.in","r",stdin);freopen("explicit.out","w",stdout)
#define pll pair <lli,lli>
#define vll vector <lli>
#define vpll vector <pll>
#define f first
#define s second
#define pb push_back
#define priq priority_queue
#define unm unordered_map
#define pf push_front
#define minii LLONG_MAX
#define all(x) x.begin(), x.end()
#define mem(a,b) memset(a,b,sizeof(a))
#define ld double
#define eb emplace_back
const long double pi = 3.1415926535898;
const lli mod=1e9+7;
//const lli mod=100000000;
using namespace std;
lli lcm (lli a, lli b) {return ((a*b)/__gcd(a,b));}
lli modpower (lli a,lli b)
{
    if(b==0) return 1;
    lli c=modpower(a,b/2)%mod;
    if (b%2) return (((a*c)%mod)*c)%mod;
    else return (c*c)%mod;
}
lli power (lli a,lli b)
{
    if(b==0) return 1;
    lli c=power(a,b/2);
    if (b%2) return (a*c*c);
    else return c*c;
}
bool prime (lli k)
{
    if (k==1||k==0) return 0;
    bool flag=1;
    lli a=sqrt(k);
    for (int i=2;i<=a;i++)
    {
        if (k%i==0)
        {flag=0;break;}
    }
    return flag;
}
lli sieve (int n)
{
    bool qrr[n+1];
    for (int i=0;i<=n;i++)
    qrr[i]=1;qrr[0]=0;qrr[1]=0;
    for (int i=2;i*i<=n;i++)
    {
        if (qrr[i]==1)
        {
            for (int j=2*i;j<=n;j+=i)
            {qrr[j]=0;}
        }
    }
    return 0;
}
int main()
{
    fio;
    lli n;
    cin>>n;
    lli a[n+1][n+1];
    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<=n;j++)
        {
            cin>>a[i][j];
        }
    }
    lli b=power(2,n);
    lli dp[b];
    mem(dp,0);
    dp[0]=1;
    for (int i=1;i<=n;i++)
    {
        for (int j=0;j<b;j++)
        {
            lli cnt=0;
            for (int k=0;k<n;k++)
            {
                if (j&(1LL<<k))
                cnt++;
            }
            if (cnt!=i-1)
            {
                continue;
            }
            for (int k=0;k<n;k++)
            {
                if (((1LL<<k)&j)==0&&a[i][k+1]==1)
                {
                    dp[j|(1LL<<k)]+=dp[j];
                    dp[j|(1LL<<k)]%=mod;
                }
            }
        }
    }
    cout<<dp[b-1];
    return 0;
}