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 Python (3.8.2) (judge verdict: Accepted)

"""
keyword: 累積和
"""

import sys
sys.setrecursionlimit(10**6)

n, k = map(int, input().split())
s = input()

nums = []
now = 1 # 今見ている数
cnt = 0 # nowがいくつ並んでいるか

for i in range(n):
    if s[i] == str(now):
        cnt += 1
    else:
        nums.append(cnt)
        now ^= 1 # 0と1を切り替える
        cnt = 1 # 新しいのをカウント
if cnt != 0:
    nums.append(cnt)

# 1-0-1-0-1-0-1って感じの配列が欲しい
# 1-0-1-0みたいに0で終わっていたら、適当に1つ足す
# 補足：1-0-1-0-1というのは、1が0個、0が2個、1が1個、、、などと並んでいるという意味。このとき1つ目の0をひっくり返すと1-1-1と連続する。1-...-1と連続している数が求める値なので、左端と右端は1にしたい
if len(nums)%2 == 0:
    nums.append(0)
    
add = 2*k+1

# 累積和を作る
tot = [0]*(len(nums)+1)
for i in range(len(nums)):
    tot[i+1] = tot[i]+nums[i]

ans = 0

# 1-0-1...の、1から始まり1で終わる範囲を見るので、偶数番目だけ見る
for i in range(0, len(nums), 2):
    
    # 次のleft, rightを計算する
    left = i
    right = min(i+add, len(nums))
    tmp = tot[right] - tot[left]
    
    ans = max(tmp, ans)
    
print(ans)