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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
 
/* attention
    long longのシフト演算には気をつけよう
    タイポした時のデバッグが死ぬほどきつくなるので変数名は最低3字くらい使った方がいいかも
    sizeは(int)とキャストしよう
    ごちゃごちゃ場合分けを考える前に全探索は考えましたか？
*/

const ll mod = 1e9+7;

// void chmod(ll &M){
//     if(M >= mod) M %= mod;
//     else if(M < 0){
//         M += (abs(M)/mod + 1)*mod;
//         M %= mod;
//     }
// }



int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n; cin >> n;
	char fi[n][n];

	for(int i=0; i<n; ++i) for(int j=0; j<n; ++j) cin >> fi[i][j];
	int ans = 0;

	for(int b=0; b<n; ++b){
		bool flg = true;
		for(int i=0; i<n; ++i){
			for(int j=0; j<n; ++j){
				if(fi[i][(j+b)%n] != fi[j][(i+b)%n]) flg = false;
				// if(b == 1){
				// 	printf("(i, j) = %c\n", fi[i][(j+b)%n]);
				// 	printf("(j, i) = %c\n", fi[j][(i+b)%n]);
				// }
			}
		}
		if(flg) ans += n;
	}

	cout << ans << endl;
	return 0;
}