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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <stdio.h>
#define NUM 10000

int main(void) {
    int i,n,ans;
    int v[NUM] = {0};
    int numodd[NUM] = {0};//奇数用の配列 
    int numeven[NUM] = {0};//偶数用の配列
    int oddmax,evenmax;

    scanf("%d",&n);

    for(i = 0; i < n; i++) {
        scanf("%d",&v[i]);//値の読み込み
    }

    for(i = 0; i < n; i+=2) {
        numodd[v[i]]++;//奇数部の数字を配列に格納　
        numeven[v[i+1]]++;//偶数部の数字を配列に格納
    }

    oddmax = numodd[0];
    evenmax = numeven[0];

    for(i = 1; i < NUM; i++) {
        if(oddmax < numodd[i]){
            oddmax = numodd[i];
        }
        if(evenmax < numeven[i]){
            evenmax = numeven[i];
        }
    }

    ans = n -(oddmax+evenmax);

    if(oddmax != evenmax){
        printf("%d\n",ans);
    }else {
        printf("%d\n",n/2);
    }
    

    return 0;
}