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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include <time.h>
typedef long long llint;
typedef long double ldouble;
// 大小
int compare( const void *a, const void *b){return *(llint *)b - *(llint *)a;}
typedef struct{
    llint aa;
    llint bb;
}frequent;
// 小大
int cmp( const void *p, const void *q ) {
    return ((frequent*)p)->bb - ((frequent*)q)->bb;
}
/*2つのうち小さい数を返す*/
llint min2(llint a,llint b){return a>=b?b:a;}


int main(void){
    llint n,i,even[100000],odd[100000],count;
    scanf("%lld",&n);
    llint num[n];
    for(i=0;i<n;i++){
        scanf("%lld",&num[i]);
    }
    for(i=0;i<100000;i++){
        even[i]=0;
        odd[i]=0;
    }
    // 0:amount 1:num[i]
    llint emax[2]={0,0},emax2[2]={0,0},omax[2]={0,0},omax2[2]={0,0};
    for(i=0;i<n;i++){
        if(i%2==0){
            even[num[i]]++;
            if(even[num[i]] > emax[0]){
                emax[0]++;
                emax[1]=num[i];
                continue;
            }
            if(even[num[i]] > emax2[0]){
                emax2[0]++;
                emax2[1]=num[i];
            }
        }
        if(i%2==1){
            odd[num[i]]++;
            if(odd[num[i]] > omax[0]){
                omax[0]++;
                omax[1]=num[i];
                continue;
            }
            if(odd[num[i]] > omax2[0]){
                omax2[0]++;
                omax2[1]=num[i];
            }
        }
    }

    if(emax[1]!=omax[1]){
        count = n - emax[0] - omax[0];
        printf("%lld\n",count);
        return 0;
    }
    count = min2(n-emax[0]-omax2[0],n-emax2[0]-omax[0]);
    printf("%lld\n",count);
    return 0;
}