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()

A = [0]
B = [0]
for s in S:
  if s == '1':
    A[-1] += 1
  elif A[-1] != 0:
    A.append(0)
  if s == '0':
    B[-1] += 1
  elif B[-1] != 0:
    B.append(0)

ans = 0
c,d,e,f = 0,0,0,0

for i in range(min(len(A),len(B))):
  a = 0
  if i == 0:
    if S[0] == '1':
      c,d = sum(A[i:min(i + K + 1,len(A))]),sum(B[i:min(i + K,len(B))])
      a = c + d
    else:
      c,d = sum(A[i:min(i + K,len(A))]) , sum(B[i:min(i + K,len(B))])
      a = c + d
  else:
    if S[0] == '1':
      if i + K <= len(A):
        c += A[min(i + K,len(A) - 1)] - A[i - 1]
        d += B[min(i + K - 1,len(B) - 1)] - B[i - 1]
        a = c + d
    else:
      if i + K <= len(A):
        c += A[min(i + K - 1,len(A) - 1)] - (0 if i == 1 else A[i - 2])
        d += B[min(i + K - 1,len(B) - 1)] - B[i - 1]
        a = c + d
  ans = max(ans,a)
print(ans)