Problem:
Score : 400 points

Problem Statement

N people are arranged in a row from left to right.

You are given a string S of length N consisting of 0 and 1, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is 0, and standing on hands if that character is 1.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

The length of the string S is N.

Each character of the string S is 0 or 1.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in PyPy2 (5.6.0) (judge verdict: Runtime Error)

N,K=map(int, raw_input().split())
S=raw_input()

if "0" not in S:
	print len(S)
	quit()

L=[]
zero_kukan=0
for i,x in enumerate(S):
	if i==0:
		cnt=1
		if x=="0": zero_kukan+=1
	else:
		if prev==x:
			cnt+=1
		else:
			L.append(cnt)
			cnt=1
			if x=="0": zero_kukan+=1
	prev=x
else:
	L.append(cnt)

K=min(K, zero_kukan)


LSUM=[0]
for x in L:
	LSUM.append(LSUM[-1]+x)


#print "L :::: ",L
#print "LSUM  :::: ",LSUM

lsum_len=len(LSUM)

#print "lsum_len  :::: ",lsum_len

h=2*K+1

ans=0
if S[0]=="1":
	for l in range(0,lsum_len-h,2):
		r=l+h
		ans=max(ans, LSUM[r]-LSUM[l])
else:
	ans=max(ans, LSUM[2*K]-LSUM[0])
	for l in range(1,lsum_len-h,2):
		r=l+h
		ans=max(ans, LSUM[r]-LSUM[l])
	
	ans=max(ans, LSUM[lsum_len-1]-LSUM[lsum_len-1-2*K])
print ans