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 Python (2.7.6) (judge verdict: Time Limit Exceeded)

exec("def nWord(): nWord.cache = raw_input() if not hasattr(nWord, 'cache') or nWord.cache=='' else nWord.cache ; pos = nWord.cache.find(nWord.sep) ; ret, nWord.cache = (nWord.cache, '') if pos==-1 else (nWord.cache[:pos], nWord.cache[pos+1:].lstrip(nWord.sep)) ; return ret") ; exec("nWord.sep = ' '") ; exec("def ___nLine2(): ret, nWord.cache = (raw_input(), nWord.cache if hasattr(nWord, 'cache') else '') if not hasattr(nWord, 'cache') or nWord.cache=='' else (nWord.cache, '') ; return ret") ; exec('def nInt(): return int(nWord())') ; exec('def nFloat(): return float(nWord())') ; exec("def ___nAuto2():\n ret = nWord()\n try: return int(ret)\n except: pass\n try: return float(ret)\n except: return ret") ; exec("def nAuto(n=None): return ___nAuto2() if n is None else [___nAuto2() for i in range(n)]") ; exec("def nLine(n=None): return ___nLine2() if n is None else [___nLine2() for i in range(n)]")

N = nAuto()
S = nAuto(N)

H = N
W = N

def fH(p, H2):
    for k in range(1, p):
        for i in range(H2):
            for j in range(W):
                if S[i][j]!=S[i+k*H2][j]:
                    return False
    return True

def fW(p, W2):
    for k in range(1, p):
        for i in range(H):
            for j in range(W2):
                if S[i][j]!=S[i][j+k*W2]:
                    return False
    return True

rate = 1
for p in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 39, 41, 43, 47]:
    while H % p == 0:
        H2 = H / p
        if fH(p, H2):
            H = H2
            rate *= p
        else:
            break
    while W % p == 0:
        W2 = W / p
        if fW(p, W2):
            W = W2
            rate *= p
        else:
            break

D = {}
for  i in range(H):
    for j in range(W):
        c = S[i][j]
        if c in D:
            D[c] += 1
        else:
            D[c] = 1

max_c = 'a'
max_count = 0
for c in D:
    if max_count < D[c]:
        max_count = D[c]
        max_c = c

L = []
for  i in range(H):
    for j in range(W):
        c = S[i][j]
        if c!=max_c:
            L.append((i, j, c))

def check(A, B):
    for i2, j2, c in L:
        i = (i2-A+N) % N
        j = (j2-B+N) % N
        if c!=S[(A+j) % N][(B+i) % N]:
            return False
    return True

ng = 0
count = 0
for A in range(H):
    for B in range(W):
        if check(A, B):
            count += 1
            if 300<=count and ng==0:
                print N*N
                exit()
        else:
            ng += 1
print count * rate