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)

#pragma GCC optimize ("O3")
#pragma GCC target ("tune=native")
#pragma GCC target ("avx")

#include <bits/stdc++.h>

// 汎用マクロ
#define ALL_OF(x) (x).begin(), (x).end()
#define REP(i,n) for (long long i=0, i##_len=(n); i<i##_len; i++)
#define RANGE(i,is,ie) for (long long i=(is), i##_end=(ie); i<=i##_end; i++)
#define DSRNG(i,is,ie) for (long long i=(is), i##_end=(ie); i>=i##_end; i--)
#define UNIQUE(v) { sort((v).begin(), (v).end()); (v).erase(unique((v).begin(), (v).end()), (v).end()); }
template<class T> bool chmax(T &a, const T &b) {if (a < b) {a = b; return true;} return false; }
template<class T> bool chmin(T &a, const T &b) {if (a > b) {a = b; return true;} return false; }
#define INF 0x7FFFFFFF
#define LINF 0x7FFFFFFFFFFFFFFFLL
#define Yes(q) ((q) ? "Yes" : "No")
#define YES(q) ((q) ? "YES" : "NO")
#define Possible(q) ((q) ? "Possible" : "Impossible")
#define POSSIBLE(q) ((q) ? "POSSIBLE" : "IMPOSSIBLE")
#define DUMP(q) cerr << "[DEBUG] " #q ": " << (q) << " at " __FILE__ ":" << __LINE__ << endl
#define DUMPALL(q) { cerr << "[DEBUG] " #q ": ["; REP(i, (q).size()) { cerr << (q)[i] << (i == i_len-1 ? "" : ", "); } cerr << "] at " __FILE__ ":" << __LINE__ << endl; }
template<class T> T gcd(const T &a, const T &b) { return a < b ? gcd(b, a) : b ? gcd(b, a % b) : a; }
template<class T> T lcm(const T &a, const T &b) { return a / gcd(a, b) * b; }

// gcc拡張マクロ
#define popcount __builtin_popcount
#define popcountll __builtin_popcountll

// エイリアス
#define DANCE_ long
#define ROBOT_ unsigned
#define HUMAN_ signed
#define CHOKUDAI_ const
using  ll = DANCE_ HUMAN_ DANCE_;
using ull = DANCE_ ROBOT_ DANCE_;
using cll = DANCE_ DANCE_ CHOKUDAI_;
using  ld = long double;
using namespace std;

// モジュール


// 処理内容
int main() {
    
    ll n; cin >> n;
    vector<ll> v(n);
    REP(i, n) cin >> v[i];

    map<ll, ll> dic0, dic1;
    REP(i, n) {
        (i%2 ? dic1 : dic0)[v[i]]++;
    }

    vector<pair<ll, ll>> d0, d1;
    for (auto &v : dic0)
        d0.emplace_back(v.second, v.first);
    for (auto &v : dic1)
        d1.emplace_back(v.second, v.first);

    sort(ALL_OF(d0), greater<pair<ll, ll>>());
    sort(ALL_OF(d1), greater<pair<ll, ll>>());


    ll coans;
    if (d0.front().second != d1.front().second) {
        coans = d0.front().first + d1.front().first;
    } else {
        d0.emplace_back(0, 0);
        d1.emplace_back(0, 0);
        coans = max(
            d0[0].first + d1[1].first,
            d0[1].first + d1[0].first
        );
    }
    cout << n - coans << endl;
    
}