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: Accepted)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <iomanip>
#include <limits>
#include <bitset>
#include <map>

using namespace std;

typedef long long ll;
typedef pair<int, int> P;
typedef pair<ll, ll> PL;
#define int ll

const ll MOD = 1000000007;
const ll INF_LL = (ll)1000000007 * 1000000007;
const int INF_INT = (int)1000000007;
const double PI = 3.14159265358979323846;

const int dx[4] = { 1, 0, -1, 0 };
const int dy[4] = { 0, 1, 0, -1 };

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define irep(itr, st) for (auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for (auto itr = (st).rbegin(); itr != (st).rend(); ++itr)

#define m0(x) memset((x), 0, sizeof((x)))
#define m1(x) memset((x), -1, sizeof((x)))

// xにはvectorなどのコンテナ
#define ALL(x) (x).begin(), (x).end()  // sortなどの引数を省略したい
#define RALL(x) (x).rbegin(), (x).rend()  // sortなどの引数を省略したい
#define SIZE(x) ((ll)(x).size())     // sizeをsize_tからllに直しておく
#define MAX(x) *max_element(ALL(x))  //最大値を求める
#define MIN(x) *min_element(ALL(x))  //最小値を求める
#define all(x) (x).begin(), (x).end()

//略記
#define pb emplace_back  // vectorヘの挿入
#define mp make_pair     // pairのコンストラクタ
#define F first          // pairの一つ目の要素
#define S second         // pairの二つ目の要素

#define BITCOUNT __builtin_popcount
#define BITCOUNT_LL(x) __builtin_popcountll(x)

#define perm(c)             \
  sort(all(c));             \
  for (bool c##p = 1; c##p; \
       c##p = next_permutation(all(c)))  //順列 123 132 213 231 312 321

#define BIT(n) (1LL << (n))

#ifdef DEBUG
#define PRINT(A) std::cout << (#A) << ": " << (A) << std::endl;
#else
#define PRINT(A)
#endif

//入力高速化
struct IoSetup {
    IoSetup() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(20);
    }
} iosetup;

template <class T>
inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template <class T>
inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}



signed main() {
    int N; cin >> N;
    map<int, int> m1,m2;
    rep(i, N) {
        int a; cin >> a;
        if (i % 2)m1[a] ++;
        else m2[a] ++;
    }
    int max1num = 0, max1val = 0;
    int max2num = 0, max2val = 0;
    vector<P> p1, p2;
    for (auto m : m1) {
        p1.push_back(make_pair(m.second, m.first));
    }
    for (auto m : m2) {
        p2.push_back(make_pair(m.second, m.first));
    }

    sort(RALL(p1));
    sort(RALL(p2));
    ll ans = N - p1[0].first - p2[0].first;
    if (p1[0].second == p2[0].second) {
        if (p1.size() == 1 && p2.size() == 1) {
            ans = N / 2;
        }
        else if (p1.size() == 1) {
            ans = N - p1[0].first - p2[1].first;
        }
        else if (p2.size() == 1) {
            ans = N - p1[1].first - p2[0].first;
        }
        else if (p1[0].first - p1[1].first > p2[0].first - p2[1].first) {
            ans = ans + (p2[0].first - p2[1].first);
        }
        else {
            ans = ans + (p1[0].first - p1[1].first);
        }
    }
    cout << ans << endl;
}