Problem:
Score : 500 points

Problem Statement

Snuke has two boards, each divided into a grid with N rows and N columns.
For both of these boards, the square at the i-th row from the top and the j-th column from the left is called Square (i,j).

There is a lowercase English letter written in each square on the first board. The letter written in Square (i,j) is S_{i,j}. On the second board, nothing is written yet.

Snuke will write letters on the second board, as follows:

First, choose two integers A and B ( 0 \leq A, B < N ).

Write one letter in each square on the second board.
Specifically, write the letter written in Square ( i+A, j+B ) on the first board into Square (i,j) on the second board.
Here, the k-th row is also represented as the (N+k)-th row, and the k-th column is also represented as the (N+k)-th column.

After this operation, the second board is called a good board when, for every i and j ( 1 \leq i, j \leq N ), the letter in Square (i,j) and the letter in Square (j,i) are equal.

Find the number of the ways to choose integers A and B ( 0 \leq A, B < N ) such that the second board is a good board.

Constraints

1 \leq N \leq 300

S_{i,j} ( 1 \leq i, j \leq N ) is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

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

Output

Print the number of the ways to choose integers A and B ( 0 \leq A, B < N ) such that the second board is a good board.

Sample Input 1

2
ab
ca

Sample Output 1

2

For each pair of A and B, the second board will look as shown below:

The second board is a good board when (A,B) = (0,1) or (A,B) = (1,0), thus the answer is 2.

Sample Input 2

4
aaaa
aaaa
aaaa
aaaa

Sample Output 2

16

Every possible choice of A and B makes the second board good.

Sample Input 3

5
abcde
fghij
klmno
pqrst
uvwxy

Sample Output 3

0

No possible choice of A and B makes the second board good.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

const int N = 3e2 + 7;

long long power[N], t[N][N], n, hsh[2][N][N][N], M = 1e9 + 7;


int main() {
	power[0] = 1;
	for (int i = 1; i < N; i++)
		power[i] = (power[i - 1] * 27) % M;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			char c;
			cin >> c;
			t[i][j] = c - 'a' + 1;
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			hsh[0][i][j][i] = hsh[1][i][j][j] = t[i][j];
			for (int k = i + 1; k <= n; k++) {
				hsh[0][i][j][k] = (hsh[0][i][j][k - 1] * 27 + t[k][j]) % M;
			}
			for (int k = j + 1; k <= n; k++) {
				hsh[1][i][j][k] = (hsh[1][i][j][k - 1] * 27 + t[i][k]) % M;
			}
		}
	}
	int res = 0;
	for (int a = 1; a < n; a++) {
		for (int b = 1; b < n; b++) {
			vector <long long> cl, rw;
			for (int p = 1; p <= n - b; p++) {
				rw.push_back((hsh[0][a + 1][b + p][n] * power[a] + hsh[0][1][b + p][a]) % M);
			}
			for (int p = 1; p <= b; p++) {
				rw.push_back((hsh[0][a + 1][p][n] * power[a] + hsh[0][1][p][a]) % M);
			}
			for (int p = 1; p <= n - a; p++) {
				cl.push_back((hsh[1][a + p][b + 1][n] * power[b] + hsh[1][a + p][1][b]) % M);
			}
			for (int p = 1; p <= a; p++) {
				cl.push_back((hsh[1][p][b + 1][n] * power[b] + hsh[1][p][1][b]) % M);
			}
			bool ok = 1;
			for (int i = 0; i < n; i++)
				ok &= (cl[i] == rw[n - i - 1]);
			res += ok;
		}
	}
	for (int a = 0; a < n; a++) {
		vector < long long > cl, rw;
		for (int p = 1; p <= n; p++)
			cl.push_back((hsh[0][a + 1][p][n] * power[a] + hsh[0][1][p][a]) % M);
		for (int p = 1; p <= n - a; p++) 
			rw.push_back(hsh[1][a + p][1][n]);
		for (int p = 1; p <= a; p++)
			rw.push_back(hsh[1][p][1][n]);
		bool ok = 1;
		for (int i = 0; i < n; i++)
			ok &= (cl[i] == rw[i]);
		res += ok;
	}
	for (int b = 0; b < n; b++) {
		vector <int> cl, rw;
		for (int p = 1; p <= n; p++) 
			rw.push_back((hsh[1][p][b + 1][n] * power[b] + hsh[1][p][1][b]) % M);
		for (int p = 1; p <= n - b; p++)
			cl.push_back(hsh[0][1][b + p][n]);
		for (int p = 1; p <= b; p++)
			cl.push_back(hsh[0][1][p][n]);
		bool ok = 1;
		for (int i = 0; i < n; i++)
			ok &= (cl[i] == rw[i]);
		if (b == 0)
			continue;
		res += ok;
	}
	cout << res;
	return 0;
}