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 _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
#define rep2(i,x,n) for(int i=x; i<(n); i++)
#define all(n) begin(n),end(n)
struct cww{cww(){ios::sync_with_stdio(false);cin.tie(0);}}star;
const long long inf = numeric_limits<long long>::max();
typedef long long ll;

ll calc(vector<ll> & vec, ll mode, ll add) {
  ll ans = 0;
  rep(i, vec.size()) {
    if (add ) {
      
    }
    if (vec[i] > 0 && mode != i) {
      ans += 1;
    }
  }
  return ans;
}

bool sortbysec(const pair<ll,ll> &a, 
              const pair<ll,ll> &b) 
{ 
    return (a.second > b.second); 
} 

vector<pair<ll, ll>> convert(vector<ll> &vec) {
  vector<pair<ll, ll>> seq;
  rep(i, vec.size()) {
    if (vec[i] == 0) {
      continue;
    }
    seq.push_back(make_pair((ll)i, vec[i]));
  }
  sort(seq.begin(), seq.end(), sortbysec);
  return seq;
}

int main()
{
  int n;
  cin >> n;
  vector<ll> vec_even(pow(10, 5)+1, 0);
  vector<ll> vec_odd(pow(10, 5)+1, 0);
  int even_size = 0;
  int odd_size = 0;
  rep(i, n) {
    int x;
    cin >> x;
    if (i%2 == 0) {
      even_size++;
      vec_even[x]++;
    } else {
      odd_size++;
      vec_odd[x]++;
    }
  }
  vector<pair<ll, ll>> even_pairs = convert(vec_even);
  vector<pair<ll, ll>> odd_pairs = convert(vec_odd);
  ll res = 0;
  if (even_pairs[0].first == odd_pairs[0].first) {
    ll a, b;
    if (even_pairs[0].second > odd_pairs[0].second) {
      a = even_size - even_pairs[0].second;
      b = odd_size - (odd_pairs.size() == 1 ? 0 : odd_pairs[1].second);
    } else {
      a = odd_size - odd_pairs[0].second;
      b = even_size - (even_pairs.size() == 1 ? 0 : even_pairs[1].second);
    }
    res = a + b;
  } else {
    ll a = even_size - even_pairs[0].second;
    ll b = odd_size - odd_pairs[0].second;
    res = a + b;
  }
  cout << res << endl;
  return 0;
}