Xorgon is an extremely delicious treat formed by the sequence $S$ of binary integers $s_1, s_2,..,s_N$. A really interesting property found in a Xorgon is that the xor of all elements in any contiguous subsequence of length $K$ in $S$ will result in $1$.   
Chef has been asked to prepare a Xorgon. However, he has at his disposal a binary sequence $X$ containing the binary integers $x_1, x_2, ...x_N$. To prepare a Xorgon, the chef may flip the value of as many binary digits in $X$ as required.(i.e. turn a $1$ to a $0$ and $0$ to a $1$). Unfortunately, flipping a digit takes a lot of time and the chef has to serve many orders. Can you help the chef calculate the minimum number of flips required to cook a Xorgon from the given $X$?

-----Input:-----
- The first line will contain two space-separated integers $N, K$.
- Next line contains N space-separated integers $x_1, x_2, ...,x_N$. 

-----Output:-----
Output in a single line minimum number of flips required to turn $X$ into a Xorgon.

-----Constraints-----
- $1 \leq K \leq N \leq 3*10^6$
- $0 \leq x_i \leq 1$

-----Sample Input:-----
7 5
1 0 0 1 1 1 1

-----Sample Output:-----
1

-----EXPLANATION:-----
Flip the last bit from 1 to 0 to obtain a Xorgon.

Difficulty: UNKNOWN_DIFFICULTY

Solutions:
from sys import stdin, stdout
a0 = 0
a1 = 1
(n, k) = stdin.readline().strip().split(' ')
(n, k) = (int(n), int(k))
arr = list(map(int, stdin.readline().strip().split(' ')))

def solve(n, k, arr):
	sol = []
	l = 0
	u = k
	while l != u:
		sol.append(arr[l:min(len(arr), u)])
		l = min(l + k, len(arr))
		u = min(u + k, len(arr))
	tiwari = []
	for i in range(k):
		titi = 0
		gao = 0
		for j in range(len(sol)):
			if len(sol[j]) > i:
				if sol[j][i] == 0:
					titi += 1
				else:
					gao += 1
		tiwari.append((titi, gao))
	minflip = (-1, -1)
	ans = 0
	ctr = 0
	for i in tiwari:
		if i[0] < i[1]:
			ans += i[0]
			ctr += (1 * a1 + a0 * a1) * a1
			if i[1] < minflip[0] or minflip[0] == -1:
				minflip = (i[1], i[0])
		else:
			ans += i[1]
			if i[0] < minflip[0] or minflip[0] == -1:
				minflip = (i[0], i[1])
	if ctr % 2 == 0:
		ans += minflip[0]
		ans -= minflip[1]
	stdout.write(str(ans) + '\n')
solve(n, k, arr)