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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

 #include <stdio.h>

char d[400][400];

int find(int x, int y, int a){
    int m,n;
    int p=1;
    for(n=1;n++-a;)for(m=0;m++-n;){
        if(d[(x+m)%a+1][(y+n)%a]==d[(x+n)%a+1][(y+m)%a])p*=1;
        else p*=0;
    }
    return p;
}

int main(){
    int a,n,m,c;
    c=0;
    scanf("%d",&a);
    for(n=0;n++-a;)scanf("%s",d[n]);
    for(n=0;n++-a;)for(m=0;m++-a;){
        c+=find(n,m,a);
    }
    printf("%d\n",c);
}