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 debug(n) cerr << #n << ':' << n << endl;
#define dline    cerr << __LINE__ << endl;

using ll = long long;
template<class T, class U> using P = pair<T,U>;
template<class T> using Heap = priority_queue<T>;
template<class T> using heaP = priority_queue<T,vector<T>,greater<T>>;
template<class T,class U> using umap = unordered_map<T,U>;
template<class T> using uset = unordered_set<T>;

template<class T>
bool ChangeMax(T&a,const T&b){
  if(a >= b) return false;
  a = b;    return true;
}

template<class T>
bool ChangeMin(T&a,const T&b){
  if(a <= b) return false;
  a = b;    return true;  
}

template<class T, size_t N, class U>
void Fill(T (&a)[N], const U&v){
    fill((U*)a,(U*)(a+N),v);
}

template<class T>
istream& operator >> (istream&is, vector<T>&v){
  for(auto&e:v)is >> e;
  return is;
}

int n;
P<int,int> Repoint(int i,int j){
  i+=n; j+=n;
  return {i%n,j%n};
}

int main(){
  cin >> n;
  vector<string> v(n);
  for(auto&e:v)cin >> e;
  int sum[300][300] = {};
  for(int i = 0; i < n; ++i){
    for(int j = 0; j < n; ++j){
      for(int k = 1; k < n; ++k){
	int x,y; tie(y,x) = Repoint(i+k,j-k);
	if(v[y][x] == v[i][j]){
	  int x1 = x-k, x2 = x-n, y1 = y, y2 = y+k-n;	  
	  tie(y1,x1) = Repoint(y1,x1);
	  tie(y2,x2) = Repoint(y2,x2);
	  sum[y1][x1]++;
	  if(k != n-1)sum[y2][x2]++;
	}
      }
    }
  }
  int ans = 0;
  int base = n * (n-1) / 2;
  for(int i = 0; i < n; ++i){
    for(int j = 0; j < n; ++j){
      if(sum[i][j] >= base) ans++;
    }
  }
  cout << ans << endl;
  return 0;
}