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 (1.15.1) (judge verdict: Accepted)

fn main() {
    let n: usize = readln();
    let s: Vec<Vec<char>> = readlns(n);
    let mut result = HashMap::new();
    let mut answer = 0;
    for a in 0..n {
        for b in 0..n {
            let key = a as isize - b as isize;
            if !result.contains_key(&key) {
                let mut ok = true;
                for i in 0..n {
                    for j in 0..n {
                        let p = (i + a) % n;
                        let q = (j + b) % n;
                        let u = (j + a) % n;
                        let v = (i + b) % n;
                        if s[p][q] != s[u][v] {
                            ok = false;
                        }
                    }
                }
                result.insert(key, if ok { 1 } else { 0 });
            }
            answer += result[&key];
        }
    }
    println!("{}", answer);
}

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{HashMap, HashSet, VecDeque};

pub trait FromLn {
    fn fromln(s: &str) -> Self;
}
pub fn readln<T: FromLn>() -> T {
    let mut buf = String::new();
    let _ = ::std::io::stdin().read_line(&mut buf).unwrap();
    T::fromln(buf.trim())
}
pub fn readlns<T: FromLn>(n: usize) -> Vec<T> {
    let mut vs = vec![];
    for _ in 0..n {
        vs.push(readln());
    }
    vs
}
macro_rules! fromln_primitives {
    ($($t:ty),*) => { $(
        impl FromLn for $t {
            fn fromln(s: &str) -> $t {
                s.parse().unwrap()
            }
        }
    )* }
}
fromln_primitives!(
    String,
    bool,
    f32,
    f64,
    isize,
    i8,
    i16,
    i32,
    i64,
    usize,
    u8,
    u16,
    u32,
    u64
);
impl<T> FromLn for Vec<T>
where
    T: FromLn,
{
    fn fromln(s: &str) -> Vec<T> {
        s.split_whitespace().map(T::fromln).collect()
    }
}
impl FromLn for Vec<char> {
    fn fromln(s: &str) -> Vec<char> {
        s.chars().collect()
    }
}
macro_rules! fromln_tuple {
    ($($t:ident),*) => {
        impl<$($t),*> FromLn for ($($t),*) where $($t: FromLn),* {
            fn fromln(s: &str) -> ($($t),*) {
                let mut it = s.split_whitespace();
                let t = ($($t::fromln(it.next().unwrap())),*);
                assert_eq!(it.next(), None);
                t
            }
        }
    }
}
fromln_tuple!(A, B);
fromln_tuple!(A, B, C);
fromln_tuple!(A, B, C, D);
fromln_tuple!(A, B, C, D, E);
fromln_tuple!(A, B, C, D, E, F);