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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

N, K = map(int, input().split())
S = list(map(int, input()))
lst = []
cnt = 0
now = 1

for i in range(N):
	if S[i] == now:
		cnt += 1
	else:
		lst.append(cnt)
		cnt = 1
		now = 1 - now
		
lst.append(cnt)

#1 0 1 0 1の形にしたい
if len(lst) % 2 == 0:
	lst.append(0)

ret = 0
for i in range(0, len(lst), 2):
	if i - 2 + 2 * K + 1 > len(lst) - 1: break
	if i == 0:
		total = sum(lst[0 : 2 * K + 1])
	else:
		total -= sum(lst[i - 2 : i])
		total += sum(lst[(i - 2) + 2 * K + 1 : min((i - 2) + 2 * K + 3, len(lst))])
	ret = max(ret, total)
	
print(ret)