output_description: Print the minimum number of elements that needs to be replaced.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      n
      v_1 v_2 ... v_n
attempt: n=int(input())
  N=list(map(int,input().split()))
  even=[]
  odd=[]
  for i in range(len(N)):
      if i%2==0:
          even.append(N[i])
      if i%2==1:
          odd.append(N[i])
  import collections
  c = collections.Counter(even)
  x=c.most_common()[0][0]
  o=even.count(x)
  d= collections.Counter(odd)
  y=d.most_common()[0][0]
  p=odd.count(y)
  s=len(even)-even.count(x)
  t=len(odd)-odd.count(y)
  if x!=y and o==len(even) and p==len(odd):
      print(0)
  elif x==y and o==len(even) and p==len(odd):
      print(n/2)
  elif x==y and (o!=len(even) or p!=len(odd)):
      if o>p:
          y=d.most_common()[1][0]
          s=len(even)-even.count(x)
          t=len(odd)-odd.count(y)
          print(s+t)
      else:
          x=c.most_common()[1][0]
          s=len(even)-even.count(x)
          t=len(odd)-odd.count(y)
          print(s+t)
  elif:
      print(s+t)
      
          
  
problem_description: Statement
  
  A sequence a_1,a_2,... ,a_n is said to be /\/\/\/ when the following
  conditions are satisfied:
  
    * For each i = 1,2,..., n-2, a_i = a_{i+2}.
    * Exactly two different numbers appear in the sequence.
  
  You are given a sequence v_1,v_2,...,v_n whose length is even. We would like
  to make this sequence /\/\/\/ by replacing some of its elements. Find the
  minimum number of elements that needs to be replaced.
samples: [{"input": "4\n    3 1 3 2", "output": "1\n    \n\nThe sequence 3,1,3,2 is not /\\/\\/\\/, but we can make it /\\/\\/\\/ by replacing\none of its elements: for example, replace the fourth element to make it\n3,1,3,1.\n\n* * *"}, {"input": "6\n    105 119 105 119 105 119", "output": "0\n    \n\nThe sequence 105,119,105,119,105,119 is /\\/\\/\\/.\n\n* * *"}, {"input": "4\n    1 1 1 1", "output": "2\n    \n\nThe elements of the sequence 1,1,1,1 are all the same, so it is not /\\/\\/\\/."}]