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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) REP(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define MSG(a) cout << #a << " " << a << endl;
#define REP(i, x, n) for (int i = x; i < n; i++)
#define OP(m) cout << m << endl
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

bool compare_by_sec(pii a, pii b)
{
    if (a.second != b.second)
        return a.second > b.second; // > ならsecondの昇順
    else
        return a.first < b.first;
}

vector<pii> even(500001), odd(500001);

int main()
{
    int n;
    cin >> n;
    int v[n];
    bool falg = 0;
    rep(i, n) cin >> v[i];

    rep(i, 50001) even[i].first = i, odd[i].first = i;

    rep(i, n)
    {
        if (i & 1)
            odd[v[i]].second++;
        else
            even[v[i]].second++;
    }

    sort(ALL(even), compare_by_sec);
    sort(ALL(odd), compare_by_sec);

    if (odd[0].first == even[0].first)
        OP(max(n - odd[0].second - even[1].second, n - odd[1].second - even[0].second));

    else
        OP(n - odd[0].second - even[0].second);

    return 0;
}