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()
ll = []
prev = ''
temp = 0
for c in s:
  t = int(c)
  if t != prev:
    prev = t
    if temp:
      ll.append(temp)
    temp = 0
  temp += 1 if t else -1
if temp:
  ll.append(temp)
# print(ll)
ans = 0
for i in range(len(ll)):
  temp = 0
  if i == 0 and ll[i] < 0:
    for j in range(min(2*k, len(ll))):
      temp += abs(ll[j])
  elif ll[i] > 0:
    for j in range(i, min(i + 2*k + 1, len(ll))):
      temp += abs(ll[j])
  ans = max(ans, temp)

print(ans)