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: Time Limit Exceeded)

n = gets.to_i
a = gets.split.map{|x| x.to_i}
h = n/2
es = Array.new(100001, 0)
os = Array.new(100001, 0)
for i in 0...h
  es[a[2*i]] += 1
  os[a[2*i+1]] += 1
end

esuf1 = esuf2 = -1
emax1 = 0
for i in 0..100000
  if es[i] == es.max
    esuf1 = i
    emax1 = es.max
    es[i] = 0
    break
  end
end
for i in 0..100000
  if es[i] == es.max
    esuf2 = i
    break
  end
end
emax2 = es.max

osuf1 = osuf2 = -1
omax1 = 0
for i in 0..100000
  if os[i] == os.max
    osuf1 = i
    omax1 = os.max
    os[i] = 0
    break
  end
end
for i in 0..100000
  if os[i] == os.max
    osuf2 = i
    break
  end
end
omax2 = os.max

if esuf1 != osuf1
  print (n - emax1 - omax1)
else
  print (n - [omax1+emax2, omax2+emax1].max)
end