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)

#include <bits/stdc++.h>
#include <random>

#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4")

#define ll long long
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define o cout<<"BUG"<<endl;
#define	IOS ios_base::sync_with_stdio(0);
#define en "\n"
#define FOR(i, j, n) for(int j = i; j < n; ++j)
#define forn(i, j, n) for(int j = i; j <= n; ++j)
#define nfor(i, j, n) for(int j = n; j >= i; --j)
#define sortv(vv) sort(vv.begin(), vv.end())
#define all(v) v.begin(), v.end()
#define ld long double
#define ull unsigned long long
#define pii pair <int, int>


using namespace std;
const ll maxn=2e5+100,inf=1e18,LOG=23,mod=1e9+7;
int block = 300, timer = 0;
const ld EPS = 1e-7;

#define bt(i) (1 << (i))
#define int ll

int n, a[maxn], cnt, ans, ans2;
int was[maxn][3];

main()
{
	cin >> n;
	forn(1, i, n)
	{
		cin >> a[i];
		was[a[i]][i % 2]++;
	}
	ans = 1e9;
	ans2 = 1e9;
	forn(1, i, 1e5)
	{
		ans = min(ans, n / 2 - was[i][0]);
	}
	forn(1, i,  1e5)
	{
		ans2 = min(ans2, n / 2 - was[i][1]);
	}
	cout << ans + ans2;
}