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;

const int mod = 1e9 + 7;

void add_self(int& a, int b) {
    a += b;
    if(a >= mod) a -= mod;
}

int main() {
    int n;
    cin>>n;
    bool a[n][n];
    for(int i=0; i<n; i++) for(int j=0; j<n; j++) cin>>a[i][j];
    vector<int> dp(1<<n, 0);
    dp[0] = 1;
    for(int mask=0; mask < (1<<n); mask++) {
        int i = __builtin_popcount(mask);
        for(int j=0; j<n; j++)
            if(a[i][j] && !(mask & (1<<j)))
                add_self(dp[mask ^ (1<<j)], dp[mask]);
    }
    cout<< dp.back() <<endl;
    return 0;
}