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)

tmp = []
while line = $stdin.gets
  tmp << line.chomp.split
end

N = tmp[0][0].to_i # 偶数
arr = tmp[1].map(&:to_i)

gusu_arr = arr.each_with_index.select{|_, i| i%2 == 0}.map{|e| e[0]}
odd_arr = arr.each_with_index.select{|_, i| i%2 != 0}.map{|e| e[0]}


h_gusu = Hash.new(0)
gusu_arr.each do |elem|
  h_gusu[elem] += 1
end
h_gusu_sorted = Hash[ h_gusu.sort_by{ |_, v| -v } ]

elm1 = h_gusu_sorted.first[0]
elm1_num = h_gusu_sorted.first[1]

h_odd = Hash.new(0)
odd_arr.each do |elem|
  h_odd[elem] += 1
end
h_odd_sorted = Hash[ h_odd.sort_by{ |_, v| -v } ]

elm2 =  h_odd_sorted.first[0]
elm2_num = h_odd_sorted.first[1]

count = 0

elm_new = -1 
h_odd_sorted.each_with_index do |obj , i|
  elm_new == obj[0] if i == 1
end

if elm1 == elm2
  if elm2_num > elm1_num
    elm1 = elm_new
    count += 1 if arr[0] != elm_new
        arr[0] = elm_new
  else
    elm2 = elm_new
    count += 1 if arr[1] != elm_new
    arr[1] = elm_new
  end
end

arr.each_with_index do |v, i|
  count += 1 if i % 2 == 0 && v != elm1
  count += 1 if i % 2 != 0 && v != elm2
end

p count