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 Rust (1.42.0) (judge verdict: Wrong Answer)

use std::collections::HashSet;
use std::io::stdin;
use std::iter::FromIterator;

fn main() {
    let mut n = String::new();
    stdin().read_line(&mut n);
    let n = n.trim().parse().unwrap();
    let mut s = vec![vec![]; n];
    for i in 0..n {
        let mut si = String::new();
        stdin().read_line(&mut si);
        s[i] = si.chars().collect();
    }
    let mut odd_ans_set = HashSet::from_iter(0..n);
    let mut even_ans_set = HashSet::from_iter(0..n);
    for i in 0..n {
        for j in 0..n {
            let mut set = HashSet::with_capacity(n);
            for d in 0..n {
                // println!("({},{}):{} ({},{}):{}", i, j, s[i][j], d, (i + j + n - d) % n, s[d][(i + j + n - d) % n]);
                if s[i][j] == s[d][(i + j + n - d) % n] {
                    // println!("({},{}),{}", d, (i + j + n - d) % n, (d + n - i + j) % n);
                    set.insert((d + n - i + j) % n);
                }
            }
            // println!("{:?}", set);
            if (i & 1) == 1 {
                odd_ans_set = HashSet::from_iter(odd_ans_set.intersection(&set).map(Clone::clone));
            } else {
                even_ans_set = HashSet::from_iter(even_ans_set.intersection(&set).map(Clone::clone));
            }
        }
    }
    // ans_set = HashSet::from_iter(ans_set.iter().map(|d| d % n));
    let ans: usize = odd_ans_set.len() * n;
    println!("{}", ans);
}