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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll m=1e9+7;
int main()
{
	ll n;
	cin>>n;
	vector<vector<ll>>arr(n,vector<ll>(n));
	for(int i=0;i<n;i++)for(int j=0;j<n;j++)cin>>arr[i][j];
	vector<ll>dp(1<<n);dp[0]=1;
		for(int mask=0;mask<(1<<n)-1;mask++)
		{
			ll a=__builtin_popcount(mask);
			for(int j=0;j<n;j++)
			{
				if(arr[a][j] && !(mask & (1<<j)))
				{
					int m1=mask|(1<<j);
					dp[m1]=(dp[m1]+dp[mask])%m;
				}
			}
		}
	cout<<dp[(1<<n)-1];
}