Problem:
Score : 1500 points

Problem Statement

There are 2N points generally positioned on the circumference of a circle, numbered 1,\dots,2N in counterclockwise order.
Here, a set of points is said to be generally positioned if, for any six distinct points U, V, W, X, Y, and Z among them, the segments UV, WX, and YZ do not intersect at the same point.
Additionally, you will be given a 2N\times 2N matrix A.

Find the number of ways to divide the 2N points into N pairs such that all of the following are satisfied:

Let us draw a red segment connecting the two points for each pair. Then, those red segments form a tree.

For each pair (P, Q), A_{P,Q} = A_{Q,P} = 1 holds.

Here, a set of segments is said to form a tree if they are all connected and form no cycles.

For example, see the figure below:

Upper left: the conditions are satisfied.

Upper right: the red segments form a cycle, so the conditions are not satisfied.

Lower left: the red segments are not connected, so the conditions are not satisfied.

Lower right: some vertices belong to no pair or multiple pairs, so the conditions are not satisfied.

Figure: A division satisfying the conditions (upper left) and divisions violating them (the others)

Notes

It can be proved that, as long as the 2N points are generally positioned, the answer does not depend on their specific positions.

Constraints

1 \leq N \leq 20

A_{i,j} is 0 or 1.

A_{i,i} is 0.

A_{i,j}=A_{j,i}

N is an integer.

Input

Input is given from Standard Input in the following format:

N
A_{1,1}...A_{1,2N}
:
A_{2N,1}...A_{2N,2N}

Output

Print the number of ways to divide the 2N points into N pairs such that all of the conditions are satisfied.
It can be proved that the answer fits into a 64-bit signed integer under the given constraints.

Sample Input 1

3
011111
101111
110111
111011
111101
111110

Sample Output 1

3

There are three possible divisions that satisfy the conditions: ((1,4),(2,6),(3,5)), ((1,3),(2,5),(4,6)), and ((1,5),(2,4),(3,6)).

Sample Input 2

4
01111100
10011111
10011100
11101111
11110111
11111011
01011101
01011110

Sample Output 2

6

Sample Input 3

8
0111101111111111
1011101111111111
1101101111011101
1110111111111111
1111011111110111
0001101111111111
1111110111011111
1111111011111111
1111111101111111
1111111110111111
1101110111011111
1111111111101111
1111011111110111
1111111111111011
1101111111111101
1111111111111110

Sample Output 3

4762

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define cs const
#define pb push_back
using namespace std;
typedef long long ll;
cs int N = 45;
int n; bool e[N][N];
ll f[N][N][N], g[N][N][N];
int main(){
	#ifdef FSYolanda
	freopen("1.in","r",stdin);
	#endif
	scanf("%d",&n); n <<= 1;
	for(int i = 1; i <= n; i++){
		char S[N]; scanf("%s", S + 1);
		for(int j = 1; j <= n; j++)
		e[i][j] = S[j] - '0';
	} 
	for(int i = 1; i < n; i++){
		f[i][i][i] = 1;
		for(int j = i + 1; j < n; j++)
		g[i][i][j] = e[i][j];
	} 
	for(int len = 3; len < n; len += 2)
	for(int l = 1, r; l <= n - len; l++){
		r = l + len - 1;
		for(int p = l; p < r; p += 2)
		for(int q = r; q > p; q -= 2){
			ll c = 0;
			for(int k = q; k <= r; k++)
			c += g[l][p][k] * f[q][r][k];
			for(int k = p + 1; k < q; k++)
			f[l][r][k] += c * f[p + 1][q - 1][k];
		}
		for(int k = l; k <= r; k++)
		for(int z = r + 1; z < n; z++)
		g[l][r][z] += f[l][r][k] * e[k][z];
	} ll ans = 0;
	for(int i = 1; i < n; i++) ans += f[1][n - 1][i] * e[n][i];
	cout << ans; return 0;
}