Let's assume that we are given a matrix *b* of size *x* × *y*, let's determine the operation of mirroring matrix *b*. The mirroring of matrix *b* is a 2*x* × *y* matrix *c* which has the following properties:
 -  the upper half of matrix *c* (rows with numbers from 1 to *x*) exactly matches *b*; -  the lower half of matrix *c* (rows with numbers from *x* + 1 to 2*x*) is symmetric to the upper one; the symmetry line is the line that separates two halves (the line that goes in the middle, between rows *x* and *x* + 1). 
Sereja has an *n* × *m* matrix *a*. He wants to find such matrix *b*, that it can be transformed into matrix *a*, if we'll perform on it several (possibly zero) mirrorings. What minimum number of rows can such matrix contain?
The first line contains two integers, *n* and *m* (1 ≤ *n*, *m* ≤ 100). Each of the next *n* lines contains *m* integers — the elements of matrix *a*. The *i*-th line contains integers *a*<sub class="lower-index">*i*1</sub>, *a*<sub class="lower-index">*i*2</sub>, ..., *a*<sub class="lower-index">*im*</sub> (0 ≤ *a*<sub class="lower-index">*ij*</sub> ≤ 1) — the *i*-th row of the matrix *a*.
n, m = map(int, input().split())
matrix = []
for i in range(n):
  matrix.append(list(map(int, input().split())))

def isequal(m1, m2):
  for i in range(len(m1)):
    for j in range(len(m1[i])):
      if m1[i][j] != m2[i][j]:
        return False
  return True

if n % 2 != 0:
  print(n)
else:
  m1 = matrix[0:int(len(matrix) / 2)]
  m2 = matrix[int(len(matrix) / 2):len(matrix)]
  m2.reverse()
  size = 0
  while isequal(m1, m2):
    size += 2
    if len(m1) % 2 != 0:
      size += len(m1)
      break
    m2 = m1[int(len(m1) / 2):len(m1)]
    m1 = m1[0:int(len(m1) / 2)]
  if size == 0:
    size = len(matrix)
  print(size)