#include <bits/stdc++.h>

#define For(i, j, k) for (int i = j; i <= k; i++)

using namespace std;

const int N = 310;

int n;
char A[N][N], B[N][N];

int main() {

	scanf("%d", &n);
	For(i, 0, n - 1) scanf("%s", A[i]);

	int ans = 0;
	For(a, 0, n - 1) {
		For(i, 0, n - 1) For(j, 0, n - 1) B[i][j] = A[(i + a) % n][j];

		bool flag = true;
		For(i, 0, n - 1) For(j, 0, n - 1) if (B[i][j] != B[j][i]) {
			flag = false;
			break;
		}
		if (flag) ans += n;
	}
	printf("%d\n", ans);

	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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.

Examples

Input

2
ab
ca


Output

2


Input

4
aaaa
aaaa
aaaa
aaaa


Output

16


Input

5
abcde
fghij
klmno
pqrst
uvwxy


Output

0

### Response

```cpp
#include <bits/stdc++.h>

#define For(i, j, k) for (int i = j; i <= k; i++)

using namespace std;

const int N = 310;

int n;
char A[N][N], B[N][N];

int main() {

	scanf("%d", &n);
	For(i, 0, n - 1) scanf("%s", A[i]);

	int ans = 0;
	For(a, 0, n - 1) {
		For(i, 0, n - 1) For(j, 0, n - 1) B[i][j] = A[(i + a) % n][j];

		bool flag = true;
		For(i, 0, n - 1) For(j, 0, n - 1) if (B[i][j] != B[j][i]) {
			flag = false;
			break;
		}
		if (flag) ans += n;
	}
	printf("%d\n", ans);

	return 0;
}
```