Codeforce Round #218 Div2 A

A. K-Periodic Array

time limit per test 1 second

memory limit per test 256 megabytes

 

This task will exclusively concentrate only on the arrays where all elements equal 1 and/or 2.

Array a is k-period if its length is divisible by k and there is such array b of length k, that a is represented by array b written exactly times consecutively. In other words, array a is k-periodic, if it has period of length k.

For example, any array is n-periodic, where n is the array length. Array [2, 1, 2, 1, 2, 1] is at the same time 2-periodic and 6-periodic and array [1, 2, 1, 1, 2, 1, 1, 2, 1] is at the same time 3-periodic and 9-periodic.

For the given array a, consisting only of numbers one and two, find the minimum number of elements to change to make the array k-periodic. If the array already is k-periodic, then the required value equals 0.

Input

The first line of the input contains a pair of integers n, k (1 ≤ k ≤ n ≤ 100), where n is the length of the array and the value n is divisible by k. The second line contains the sequence of elements of the given array a₁, a₂, ..., a_n (1 ≤ a_i ≤ 2), a_i is the i-th element of the array.

Output

Print the minimum number of array elements we need to change to make the array k-periodic. If the array already is k-periodic, then print 0.

Sample test(s)

Input

6 2 
2 1 2 2 2 1

Output

1

Input

8 4 
1 1 2 1 1 1 2 1

Output

0

Input

9 3 
2 1 1 1 2 1 1 1 2

Output

3

Note

In the first sample it is enough to change the fourth element from 2 to 1, then the array changes to [2, 1, 2, 1, 2, 1].

In the second sample, the given array already is 4-periodic.

In the third sample it is enough to replace each occurrence of number two by number one. In this case the array will look as [1, 1, 1, 1, 1, 1, 1, 1, 1] — this array is simultaneously 1-, 3- and 9-periodic.

#include <cstdio>
#include <cmath>
#include <vector>
#include <stack>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 1000
#define INF 0x7fffffff
int n, m, k, x, y, d,t;
int a[maxn][maxn];
int gcd(int n, int m){
    if (m == 0)return n;
    return gcd(m, n%m);
}
int main(){
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n/m; i++){
        for (int j = 0; j < m; j++)
            scanf("%d", &a[i][j]);
    }
    int s = 0;
    for (int i = 0; i < m; i++){
        x = 0; y = 0;
        for (int j = 0; j < n/m;j++)
        if (a[j][i] == 1)x++;
        else y++;
        s += min(x,y);
    }
    printf("%d\n", s);
    return 0;
}