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 = map(int, input().split())
s = list(map(int, input()))
count_1 = 0


def re_cum_list(li):
  count_0 = 0
  count_1 = 0
  re_li = []
  for i in li:
    if i == 0:
      count_0 -= 1
      if count_1 > 0:
        re_li.append(count_1)
        count_1 = 0
    else:
      count_1 += 1
      if count_0 < 0:
        re_li.append(count_0)
        count_0 = 0
  if count_1 > 0:
    re_li.append(count_1)
  if count_0 < 0:
    re_li.append(count_0)
  return re_li


cum_1_s = re_cum_list(s)

sum_list = []
su = 0
count = 0
j = 0
while j <= len(cum_1_s):
  for i in range(j, len(cum_1_s)):
    if cum_1_s[i] > 0:
      pass
    else:
      if count >= k:
        break
      count += 1
    su += abs(cum_1_s[i])
  sum_list.append(su)
  j += 1
  count = 0
  su = 0


print(max(sum_list))