#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long int lli;
lli dp[1<<21];
int main()
{
    ios::sync_with_stdio(0); 
    cin.tie(0); cout.tie(0);
    lli n;
    cin>>n;
    lli i,j,a[n][n],M=1e9+7;
    
    for(i=0;i<n;i++) for(j=0;j<n;j++) cin>>a[i][j];
    
    dp[0]=1;
    for(i=1;i<1<<n;i++)
    {
        for(j=0;j<n;j++)
        {
            if(i&(1<<j) && a[__builtin_popcount(i)-1][j])
                dp[i]=(dp[i]+dp[i^(1<<j)])%M;
        }
    }
    
    cout<<dp[(1<<n)-1];
    return 0;
}
### Prompt

In Cpp, your task is to solve the following problem:

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.

Examples

Input

3
0 1 1
1 0 1
1 1 1


Output

3


Input

4
0 1 0 0
0 0 0 1
1 0 0 0
0 0 1 0


Output

1


Input

1
0


Output

0


Input

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


Output

102515160

### Response

```cpp
#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long int lli;
lli dp[1<<21];
int main()
{
    ios::sync_with_stdio(0); 
    cin.tie(0); cout.tie(0);
    lli n;
    cin>>n;
    lli i,j,a[n][n],M=1e9+7;
    
    for(i=0;i<n;i++) for(j=0;j<n;j++) cin>>a[i][j];
    
    dp[0]=1;
    for(i=1;i<1<<n;i++)
    {
        for(j=0;j<n;j++)
        {
            if(i&(1<<j) && a[__builtin_popcount(i)-1][j])
                dp[i]=(dp[i]+dp[i^(1<<j)])%M;
        }
    }
    
    cout<<dp[(1<<n)-1];
    return 0;
}
```