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
v=gets.chomp.split(" ").map { |e| e.to_i }
even = []
odd = []

for i in 0..n-1 do

  if i%2 == 0
    even[i/2] = v[i]
  else
    odd[i/2] = v[i]
  end
end

even = even.sort
odd = odd.sort

half = n/2
dp_even=0
dp_odd=0
max_even = 0
max_odd = 0
indx_even = 0
indx_odd = 0
string_even = even[0]
string_odd = odd[0]

for i in 0..half-1 do
  if even[i] != even[indx_even]
    indx_even = i
    if dp_even > max_even
      string_even = even[i]
      max_even = dp_even
    end
    dp_even = 0
  end


  if even[i] == even[indx_even]
    dp_even += 1
  end

end

if dp_even > max_even
  string_even = even[half-1]
  max_even = dp_even
end


for i in 0..half-1 do
  if odd[i] != odd[indx_odd]
    indx_odd = i
    if dp_odd > max_odd && string_even != odd[i-1]
      max_odd = dp_odd
    end
    dp_odd = 0
  end

  if odd[i] == odd[indx_odd]
    dp_odd += 1
  end
end


if dp_odd > max_odd && string_even != string_odd
  max_odd = dp_odd
end


puts (half - max_even) + (half - max_odd)