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 Python (2.7.6) (judge verdict: Runtime Error)

import collections

N = int(raw_input().replace("\n", ""))

a_list = map(int, raw_input().replace("\n", "").split(" "))
even_list = a_list[::2]
odd_list = a_list[1::2]

len_vec = N / 2

even_c = collections.Counter(even_list)
odd_c = collections.Counter(odd_list)



even_first_value, even_first_count = even_c.most_common()[0]
odd_first_value, odd_first_count  = odd_c.most_common()[0]
#(even_first_value, even_first_count), (even_second_value, even_second_count) = even_c.most_common()[:2]
#(odd_first_value, odd_first_count), (odd_second_value, odd_second_count) = odd_c.most_common()[:2]

if even_first_value == odd_first_value and len(even_c) == 1 and len(odd_c) == 1:
    cost = len_vec
elif even_first_value != odd_first_value:
    cost =  (even_first_count + odd_first_count)
elif len(even_c) == 1 and len(odd_c) > 1:
    odd_second_value, odd_second_count = odd_c.most_common(1)
    cost = len_vec + odd_second_count
elif len(odd_c) == 1 and len(even_c) > 1:
    even_second_value, even_second_count = even_c.most_common(1)
    #cost = len_vec + odd_second_count
    cost = len_vec + even_second_count
else: # ryouhou no first ga onaji de hutatuijou arutoki
    odd_second_value, odd_second_count = odd_c.most_common(1)
    even_second_value, even_second_count = even_c.most_common(1)
    cost = max(even_first_count+odd_second_count, even_second_count+odd_first_count)

print 2* len_vec - cost