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: Runtime Error)


(N, K) = map(int, input().split(" "))
S = input()

nowOn = S[0]
count = 0
serialList = []

for i in S:

    if(nowOn == i):
        count += 1

    else:
        serialList.append(count)
        count = 1
        nowOn = i

serialList.append(count)

T = 2 * K + 1
A = 0

#serialList -> ["1", "0", "1", "0", ...]
if S[0] == "0":
    serialList.insert(0, 0)
    
#exception
# --???

maxA = 0
for i in range(T):
    A += serialList[i]
    maxA = A
    
for i in range(T, len(serialList), 2):
    A -= serialList[i - T]
    A -= serialList[i - T + 1]
    A += serialList[i]

    if(i+1 < len(serialList)):
        A += serialList[i + 1]

    if maxA < A:
        maxA = A
        
print(maxA)