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 OCaml (4.02.3) (judge verdict: Accepted)

let rec for_fold a b v f = if a >= b then v else for_fold (a+1) b (f v a) f
let () =
  Scanf.scanf "%d" @@ fun n ->
  let s = Array.init n (fun _ -> Scanf.scanf " %s" (fun x -> x)) in
  let s = Array.init (2*n-1) (fun i ->
    Array.init n (fun j -> s.(i mod n).[j mod n])) in
  for_fold 0 n 0 (fun c i ->
    let sym =
      for_fold 0 (n*n) true (fun b j ->
        b && s.(i + j/n).(j mod n) = s.(i + j mod n).(j/n))
    in c + if sym then n else 0)
  |> Printf.printf "%d\n"