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>

int main()
{
    int in, n[3], out;
    int i;


    scanf("%d", &in);
    
    n[0] = in / 100;
    n[1] = in / 10 - n[0]*10;
    n[2] = in - (n[0]*100 + n[1]*10);

//printf("%d %d %d\n", n[0], n[1], n[2]);

    if((n[0] >= n[1] && n[0] > n[2]))
    out = n[0] *100 + n[0]*10+n[0];
    else if(n[0] < n[1] || n[0] < n[2])
    out = (n[0]+1)*100 + (n[0]+1)*10+(n[0]+1);
    else if (n[0] == n[1] && n[0] == n[2])
    out = n[0]*100+n[1]*10+n[0];
    else
    out = n[0]*100+n[1]*10+n[0];
    printf("%d\n", out);
    return 0;

}