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)

n = gets.chomp.to_i
ary = gets.split(" ").map(&:to_i)

odd = ary.each_slice(2).map(&:first)
even = ary.each_slice(2).map(&:last)

if ary.uniq.size == 1
  puts n/2
  exit
end

if odd.uniq.size == 1 && even.uniq.size == 1
  puts 0
  exit
end

m = 0
max = 0

if odd.uniq.size == 1
  m = even.uniq
  m.size do |i|
    if even.include(m[i]) > max && odd[0] != m[i]
      max = m[i]
    end
  end
  puts even.size - max
elsif even.uniq.size == 1
  m = odd.uniq
  m.size do |i|
    if odd.include(m[i]) > max && even[0] != m[i]
      max = m[i]
    end
  end
  puts odd.size - max
else
  if odd.max_by {|x| odd.count(x)} != even.max_by {|x| even.count(x)}
    puts add.size - add.count(odd.max_by {|x| odd.count(x)}) + 
    (even.size - enen.count(even.max_by {|x| even.count(x)}))
  end
end