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.
[n,k] = list(map(int, input().split()))
a = list(map(int, input().split()))

def abs_diff(ls1, ls2):
	out_val =0
	for i, j in zip(ls1, ls2):
		out_val += abs(i-j)
	
	return out_val

dt ={}
ls0 =[]
lsout = 0

for i in range(1,int(n/k)+1):
	dt[f"{a[k*(i-1):k*i]}"] = 0
	ls0.append(a[k*(i-1):k*i])

for i in range(1,int(n/k)+1):
	dt[f"{a[k*(i-1):k*i]}"] +=1

val = max(dt.values())
if val == 1:
	if sum(a) > 3*n:
		mode = [2 for _ in range(k)]
	else:
		mode = [1 for _ in range(k)]
	for i in ls0:
		lsout += abs_diff(i, mode)

else:
	for key, value in dt.items():
		if value == val:
			mode = list(map(int, key[1:-1].split(',')))

		for i in ls0:
			lsout += abs_diff(i, mode)

print(lsout)