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: Runtime Error)

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

def runLength(s):
  if len(s) == 0: return []
  if len(s) == 1: return [(s, 1)]
  res = []
  lc = 1
  for i in range(1, len(s)):
    if s[i-1] == s[i]:
      lc += 1
    else:
      res.append((s[i-1], lc))
      lc = 1
  res.append((s[i], lc))
  return res

S = runLength(input())
accsum = [0] * (1 + (S[0][0] == "0"))
for i in range(len(S)):
  accsum.append(accsum[-1] + S[i][1])
if S[-1][0] == "0":
  accsum.append(accsum[-1])
print(max([accsum[i] - accsum[i-2*K-1] for i in range(2*K+1, len(accsum), 2)]) if N >= 2*K else N)