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: Time Limit Exceeded)

N, K = list(map(int, input().split()))
S = input()

def search(first, count):
  change = 0
  past = "1"
  for i in range(first, len(S)):
    now = S[i]
    if past != now:
      change += 1
      if change == count*2:
        #print(S[first], S[i])
        for j in range(i, len(S)):
          if "1" != S[j]:
            #print(j, first)
            return j-first
    past = S[i]
  #print(len(S), first)
  return len(S)-first

max_len = 0
past = ""
for i in range(len(S)):
  now = S[i]
  if past != now:
    #print(i, S[i], search(i, K))
    max_len = max([search(i, K), max_len])
  past = S[i] 
  

#print(N, K, S)
print(max_len)