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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"
using namespace std;
#define ll long long int
#define rep(i,n) for( int i = 0; i < n; i++ )
#define rrep(i,n) for( int i = n; i >= 0; i-- )
#define REP(i,s,t) for( int i = s; i <= t; i++ )
#define RREP(i,s,t) for( int i = s; i >= t; i-- )
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define INF 2000000000
#define mod 1000000007
#define INF2 1000000000000000000


int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
	int N; cin >> N;
	string S[N * 2];
	rep(i, N) cin >> S[i];
	rep(i, N) S[i] += S[i];
	rep(i, N) S[i].pop_back();
	rep(i, N) S[N + i] = S[i];

	int flag1 = 1;
	int flag2 = 1;
	int flag3 = 1;
	rep(i, N) {
		rep(j, N) {
			if (S[i][j] != S[j][i]) {
				flag1 = 0;
				break;
			}
		}
	}
	rep(i, N - 1) {
		rep(j, N) {
			// cout << S[i + 1][j] << " " << S[0][(j + i) % N];
			if (S[i + 1][j] != S[0][(j + i + 1) % N]) {
				flag3 = 0;
				break;
			}
		}
		// cout << endl;
	}
	rep(i, N) {
		rep(j, N) {
			if (S[i][j] != S[N - j - 1][N - i - 1]) {
				flag2 = 0;
				break;
			}
		}
	}
	int flag4 = 0;
	rep(A, N) {
		int ok = 1;
		rep(i, N) {
			rep(j, N) {
				if (S[i + A][j] != S[j + A][i]) {
					ok = 0;
					break;
				}
				if (!ok) break;
			}
		}
		flag4 += ok;
	}
	/* 
	rep(B, N) {
		int ok = 1;
		rep(i, N) {
			rep(j, N) {
				if (S[i][j + B] != S[j][i + B]) {
					ok = 0;
					break;
				}
				if (!ok) break;
			}
		}
		flag4 += ok;
	}
	*/
	//dump(flag4);
	if (flag3 || (flag4 == N)) cout << N * N << endl;
	else if (flag4) cout << N << endl;
	else cout << 0 << endl;

    return 0;
}