Problem:
Score : 300 points

Problem 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.

Constraints

2 \leq n \leq 10^5

n is even.

1 \leq v_i \leq 10^5

v_i is an integer.

Input

Input is given from Standard Input in the following format:

n
v_1 v_2 ... v_n

Output

Print the minimum number of elements that needs to be replaced.

Sample Input 1

4
3 1 3 2

Sample Output 1

1

The sequence 3,1,3,2 is not /\/\/\/, but we can make it /\/\/\/ by replacing one of its elements: for example, replace the fourth element to make it 3,1,3,1.

Sample Input 2

6
105 119 105 119 105 119

Sample Output 2

0

The sequence 105,119,105,119,105,119 is /\/\/\/.

Sample Input 3

4
1 1 1 1

Sample Output 3

2

The elements of the sequence 1,1,1,1 are all the same, so it is not /\/\/\/.

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

#方針: 最頻値に揃える
n = gets.to_i
v = gets.split
even = v.select.with_index{|v, i| i.even?}
odd = v.select.with_index{|v, i| i.odd?}

odd2 = odd.group_by{|v| v}.sort_by{|key, value| value.size}.reverse
even2 = even.group_by{|v| v}.sort_by{|key, value| value.size}.reverse

#最頻値が異なるなら素直に出力
if odd2[0]!=even2[0]
  p n/2-odd2[0][1].size + n/2-even2[0][1].size
  p "hoge"

else
#最頻値が同じなら二番目に多いやつに切り替え
  ans1, ans2 = n/2, n/2
  if odd2[1]
    ans1 = n/2-odd2[1][1].size + n/2-even2[0][1].size
  end

  if even2[1]
    ans2 = n/2-odd2[0][1].size + n/2-even2[1][1].size
  end
  
  puts [ans1,ans2].min
end