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.




     
    

def solve():
    
    n_comprimento_array,k = map(int,input().split())
    lista_matriz = list(map(int,input().split()))
    sol = 0
    incremento_a  = 0
    while incremento_a < k:
        list_aux = []
        incremento_b = incremento_a

        while(incremento_b < n_comprimento_array):
            list_aux.append(lista_matriz[incremento_b])
            incremento_b = incremento_b + k

        count_one = list_aux.count(1)
        count_two = list_aux.count(2)
        valor_min = min(count_one,count_two)
        sol = sol + valor_min
        incremento_a = incremento_a+1
    print(sol)

solve()




		 					  			 			    	  	 				