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)

import itertools

def getint(): return int(input())
def getints(): return list(map(int, input().split()))

n,k=getints()
s=input()
n=len(s)

rl = []
now, cnt = '0', 0

for i in range(n):
    if now != s[i]:
        rl.append(cnt)
        now = s[i]
        cnt = 1
    else:
        cnt += 1
if cnt > 0:
    rl.append(cnt)
acc=[0]*len(rl)
acc[0]=rl[0]
for i in range(1, len(rl)):
    acc[i] = acc[i-1]+rl[i]
res = 0
for i in range(len(acc)):
    if i % 2 == 0:
        i0 = i - 2
        j  = i + 2 * k - 1
    else:
        i0 = i - 1
        j  = i + 2 * k
    tmp1 = acc[i0] if i0 >= 0      else 0
    tmp2 = acc[j]  if j < len(acc) else 0
    res = max(res, tmp2-tmp1)


print(res)