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 (3.4.3) (judge verdict: Runtime Error)

n = int(input())
v_odd = {}
v_even = {}
vlist = list(map(int, input().split()))
for i in range(n):
    if i % 2 == 1:
        if vlist[i] in v_odd:
            v_odd[vlist[i]] += 1
        else:
            v_odd[vlist[i]] = 1
    else:
        if i in v_even:
            v_even[vlist[i]] += 1
        else:
            v_even[vlist[i]] = 1
omax = max(v_odd, key=(lambda x: v_odd[x]))
emax = max(v_even, key=(lambda x: v_even[x]))
if omax != emax:
    orewrite = sum(v_odd[i] for i in v_odd) - v_odd[omax]
    erewrite = sum(v_even[i] for i in v_even) - v_even[emax]
    ans = orewrite + erewrite
    print(ans)
else:
    lo = len(v_odd)
    le = len(v_even)
    if lo == 1 and le == 1:
        print(v_odd[omax])
    elif lo != 1:
        sum = sum(v_odd[i] for i in v_odd)
        v_odd[omax] = 0
        osecmax = max(v_odd, key=(lambda x: v_odd[x]))
        ans = sum - osecmax
        print(ans)
    else:
        sum = sum(v_even[i] for i in v_even)
        v_even[emax] = 0
        esecmax = max(v_even, key=(lambda x: v_even[x]))
        ans = sum - esecmax
        print(ans)