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: Wrong Answer)

n,k=map(int,input().split())
s=input()
a0=[]
a1=[]
count=0
if s[0]=='0':
  tmp='0'
else:
  tmp='1'
for i in range(len(s)):
  if s[i]==tmp:
    count+=1
  else:
    if tmp=='0':
      a0.append(count)
      count=1
      tmp='1'
    else:
      a1.append(count)
      count=1
      tmp='0'
if tmp=='0':
  a0.append(count)
else:
  a1.append(count)
if len(a0)<=k:
  print(n)
else:
  ans=0
  if s[0]=='0':
    for i in range(len(a0)-k+1):
      tmp=sum(a0[i:i+k])+sum(a1[i-1:i+k])
      if ans<tmp:
        ans=tmp
  else:
    for i in range(len(a0)-k+1):
      tmp=sum(a0[i:i+k])+sum(a1[i:i+k+1])
      if ans<tmp:
        ans=tmp
  print(ans)