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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
//#include <boost/multiprecision/cpp_int.hpp>
//typedef boost::multiprecision::cpp_int ll;
typedef long double dd;
#define i_7 (ll)(1E9+7)
//#define i_7 998244353
#define i_5 i_7-2
ll mod(ll a){
    ll c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}
typedef pair<ll,ll> l_l;
ll inf=(ll)1E17;
#define rep(i,l,r) for(ll i=l;i<=r;i++)
#define pb push_back
ll max(ll a,ll b){if(a<b)return b;else return a;}
ll min(ll a,ll b){if(a>b)return b;else return a;}
void Max(ll &pos,ll val){pos=max(pos,val);}//Max(dp[n],dp[n-1]);
void Min(ll &pos,ll val){pos=min(pos,val);}
void Add(ll &pos,ll val){pos=mod(pos+val);}
dd EPS=1E-9;
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
///////////////////////////////////////


#define N 100005

int main(){fastio
    ll n;cin>>n;
    n=n/2;
    ll a[N],b[N];memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    rep(i,0,n-1){
        ll x,y;cin>>x>>y;
        a[x]++;b[y]++;
    }
    ll ma=-1,mapos=-1;
    ll ma2=-1,ma2pos=-1;
    rep(i,0,N-1){
        if(a[i]>ma){
            ma2=ma;
            ma2pos=mapos;
            ma=a[i];
            mapos=i;
        }else if(a[i]>ma2){
            ma2=a[i];
            ma2pos=i;
        }
    }
    ll ans=0;
    rep(i,0,N-1){
        if(i==mapos){
            Max(ans,ma2+b[i]);
        }else{
            Max(ans,ma+b[i]);
        }
    }
    cout<<n*2-ans<<endl;
    
    
    return 0;
}