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.
import math
n,k=map(int,input().split())
l=list(map(int,input().split()))

def helper(i,arr):
    if arr[:k]*(n//k)==arr:
        return 0
        
    if i>=len(arr):
        return math.inf
        
    if arr[i]==1:
        op1=1+helper(i+1,arr[:i]+[2]+arr[i+1:])
    if arr[i]==2:
        op1=1+helper(i+1,arr[:i]+[1]+arr[i+1:])
        
    op2=helper(i+1,arr)
    return min(op1,op2)
    
print(helper(0,l))