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)

def getN():
    return int(input())
def getNM():
    return map(int, input().split())
def getList():
    return list(map(int, input().split()))
def getArray(intn):
    return [int(input()) for i in range(intn)]
def input():
    return sys.stdin.readline().rstrip()

from collections import defaultdict, deque, Counter
from sys import exit
import heapq
import math
import copy
from operator import mul
from functools import reduce
from bisect import bisect_left, bisect_right

import sys
sys.setrecursionlimit(1000000000)

N, K = getNM()
S = input()

flag = S[0]
sta = 1
lista = []
# 14 2
# 11101010110011なら
# ２回まで0の箇所を消すことができる

# 00010を['0', 1, 3], ['1', 4, 4], ['0', 5, 5]という風に整頓する
for i in range(1, N):
    if flag != S[i]:
        # flag(0か1か), sta(どこから続いているか), i(最終的にどこまで続いたか)
        lista.append([flag, sta, i])
        flag = S[i]
        sta = i + 1
lista.append([flag, sta, N])
ans = 0
np = len(lista)
for i in range(np):
    if lista[i][0] == '0':
        index = min(i + 2 * K - 1, np - 1)
        opt = lista[index][2] - lista[i][1] + 1
        ans = max(ans, opt)
    else:
        index = min(i + 2 * K, np - 1)
        opt = lista[index][2] - lista[i][1] + 1
        ans = max(ans, opt)
print(ans)