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: Time Limit Exceeded)

#[allow(unused_macros)]
macro_rules! input_byline {
    ($($r:tt)* ) => {
        let stdin = std::io::stdin();
        let mut q = std::collections::VecDeque::new();
        let mut next = || {
            if q.is_empty() {
                let mut s = String::new();
                stdin.read_line(&mut s);
                for v in s.split_whitespace() {
                    q.push_back(v.to_string());
                }
            }
            q.pop_front().unwrap()
        };
        input_inner!{next, $($r)*}
    };

}

#[allow(unused_macros)]
macro_rules! input {
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

trait Printable {
    fn p(&self);
}

impl<T: std::fmt::Display> Printable for T {
    fn p(&self) {
        println!("{}", self);
    }
}

impl<T: std::fmt::Display> Printable for [T] {
    fn p(&self) {
        for i in 0..self.len() {
            if i != 0 {
                print!(" ");
            }
            print!("{}", self[i]);
        }
        println!("");
    }
}

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

#[allow(dead_code)]
type Graph = Vec<Vec<usize>>;

struct RollingHash {
    modp: Vec<usize>,
    base: Vec<usize>,
    hash: Vec<Vec<usize>>,
    pow: Vec<Vec<usize>>,
}

impl RollingHash {
    fn new(s: &Vec<char>) -> RollingHash {
        let n = s.len();
        let modp = vec![1000000007, 1000000009];
        let base = vec![1009, 1007];
        let mut hash = vec![vec![0; 2]; n + 1];
        let mut pow = vec![vec![1; 2]; n + 1];

        for i in 0..n {
            for j in 0..2 {
                hash[i + 1][j] = ((hash[i][j] + s[i] as usize) * base[j]) % modp[j];
                pow[i + 1][j] = (pow[i][j] * base[j]) % modp[j];
            }
        }

        RollingHash {
            modp: modp,
            base: base,
            hash: hash,
            pow: pow,
        }
    }

    // [l..r)
    fn get(&self, l: usize, r: usize) -> (usize, usize) {
        let mut res = vec![0; 2];
        for i in 0..2 {
            res[i] = (self.hash[r][i] + self.modp[i]
                - (self.hash[l][i] * self.pow[r - l][i]) % self.modp[i])
                % self.modp[i];
        }
        (res[0], res[1])
    }
}

fn main() {
    input! {
        n: usize,
        v: [chars; n],
    }
    let mut row = vec![];
    let mut col = vec![];
    for i in 0..n {
        let mut vr = vec![];
        for j in 0..n * 2 {
            vr.push(v[i][j % n]);
        }
        row.push(RollingHash::new(&vr));
    }
    for i in 0..n {
        let mut vc = vec![];
        for j in 0..n * 2 {
            vc.push(v[j % n][i]);
        }
        col.push(RollingHash::new(&vc));
    }

    let mut ans = 0;
    for i in 0..n {
        for j in 0..n {
            let mut ok = true;
            for k in 0..n {
                ok &= row[(i + k) % n].get(j, j + n) == col[(j + k) % n].get(i, i + n);
                if !ok {
                    break;
                }
            }
            if ok {
                ans += 1;
            }
        }
    }
    ans.p();
}