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 _count(list, item):
	count = 0
	for c in list:
		if c == item:
			count += 1
	return count
 
def common(list):
	counts = []
	maxcount = 0
	for stuff in list:
		if _count(list, stuff) > maxcount:
			maxitem = stuff
			maxcount = _count(list, stuff)
	return maxitem


aa = input().split()
n = int(aa[0])
k = int(aa[1])
a = input().split()


matrix = []
new = []
for i in range(1, n + 1):
	if i % k == 0:
		new.append(a[i - 1])
		matrix.append(new)
		new = []
	else:
		new.append(a[i-1])

colums = []
answer = 0
for j in range(len(matrix[0])):
	colum = []
	for i in range(len(matrix)):
		colum.append(matrix[i][j])
	colums.append(colum)

for rows in matrix:
	for i in range(len(rows)):
		if rows[i] != common(colums[i]):
			answer += 1
print(answer)