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.
# cook your dish here
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])
 #print(ans,ctr)
 #print(tiwari)
 #print(minflip)
 if ctr%2==0:
   ans+=minflip[0]
   ans-=minflip[1]


 stdout.write(str(ans)+"\n")

solve(n,k,arr)
# cook your dish here
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])
 #print(ans,ctr)
 #print(tiwari)
 #print(minflip)
 if ctr%2==0:
   ans+=minflip[0]
   ans-=minflip[1]


 stdout.write(str(ans)+"\n")

solve(n,k,arr)