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 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e8f8616e518b67af73332593c4c73ae9f207dc04.png" style="max-width: 100.0%;max-height: 100.0%;"/> 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.
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 *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 2), *a*<sub class="lower-index">*i*</sub> is the *i*-th element of the array.
#include<bits/stdc++.h>
#define  mem(a)          memset(a,-1,sizeof(a))
#define  all(a)          (a.begin(),a.end())
#define  clr(a)          memset(a,0,sizeof(a))
#define   pb             push_back
#define   PI             acos(-1.0)
#define   max3(a,b,c)    max(a,max(b,c))
#define   min3(a,b,c)    min(a,min(b,c))
#define   read           freopen("input.txt", "r", stdin)
#define   write          freopen("output.txt", "w", stdout)
#define   ll             long long
#define    s1(a)         scanf("%d",&a)
#define   s2(a,b)        scanf("%d %d",&a,&b)
#define   cp             printf("Case %d: ",kk++)
#define   mod            1000000007
#define   pf1(a)         printf("%d",a)
#define   pf2(a,b)       printf("%d %d",a,b)
#define For(i,a,b)       for(i=a;i<b;i++)
#define gofast           ios_base::sync_with_stdio(false);cin.tie(0);
using namespace std;

int main(){

int n,k,a[100000];
cin>>n>>k;
for(int i =0;i<n;i++){
	cin>>a[i];
}
int sum = 0;
for(int i =0;i<k;i++){
	int sum2 = 0;
     for(int j = i;j<n;j += k){
     	sum2 += (a[j]-1);
     	// cout<<sum2<<endl;
     }
     if(sum2<(n/k- sum2)){
         sum += sum2;
     }
     else{
     	sum += n/k-sum2;
     }
   

}
cout<<sum<<endl;

}
	       	 				  				  					