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: Accepted)

N, K = map(int, input().split())
S = input()
block = []
prev = 1
cnt = 0

for s in S:
    if str(prev) != s:
        prev = 1 - prev
        block.append(cnt)
        cnt = 1
    else:
        cnt += 1
else:
    if cnt != 0:
        block.append(cnt)
if len(block) %2 == 0:
    block.append(0)

length = 2 * K + 1
left = 0
right = min(length, len(block))
tmp = sum(block[left:right])
ans = tmp

for i in range(2,len(block),2):
    new_left = left + 2
    tmp -= sum(block[left:new_left])
    left = new_left
    new_right = min(right+2, len(block))
    tmp += sum(block[right:new_right])
    right = new_right
    ans = max(tmp, ans)
print(ans)