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)

a = []

n,k = map(int, input().split())
for x in range(k+1):
  a.append(0)
q = input()

p = 0
c = 0
mx = 0

for i in range(n-1):
  p = p+1
  c = c+1  
  if q[i] == q[i+1]:
    lalala = 1

  elif (int(q[i]) == 1) & (int(q[i+1]) == 0) :
    
    p = p-a[0]
    if p > mx:
      mx = p
    
  elif (int(q[i]) == 0) & (int(q[i+1]) == 1):
    a.append(c)
    a.pop(0)
    c = 0

if int(q[n-1]) == 1:
  if p+1> mx:
    mx = p +1
else:
  if p+1 - a[0] > mx:
    mx = p - a[0] +1
  
  


print(mx)