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 (3.4.3) (judge verdict: Accepted)

import sys
n, k = map(int, input().split())
s = input()
sList = [] #連続する1の数、0の数、1の数...と格納するリスト
switch = int(s[0])
num = 1
if int(s[0]) == 0:
  sList.append(0)
if n > 1:
  for i in range(1, n):
    if int(s[i]) == switch:
      num += 1
    else:
      sList.append(num)
      num = 1
      switch = int(s[i])
sList.append(num)
if len(sList) // 2 < k: #全部ひっくり返せる
  print(n)
  sys.exit()
elif len(sList) % 2 == 0: #最後が0の数で終わっている場合、1の数で終わらせる
  sList.append(0)
  
s = sum(sList[0:2*k+1])
sMax = s

for i in range(1, (len(sList)-2*k) // 2 + 1): #リストの2k+1個の部分和のうち、最大のものが答え
  s += - sList[2*i-2] - sList[2*i-1] + sList[2*i+2*k] + sList[2*i+2*k-1]
  if sMax < s:
    sMax = s
    
print(sMax)