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)

n, k = map(int, input().split())
s = list(input())

seq = []

if s[0] == '1':
    seq.append(0)

prev = s[0]
i = 0
"""
while i < n:
    f = True
    for j in range(i+1, n):
        if s[j] != prev:
            seq.append(j-i)
            f = False
            prev = s[j]
            break
    if f:
        i += 1
    else:
        i = j
"""
# 1回のforでseqを取得する
prev = s[0]
seq_val = 1
for i in range(1, n):
    if s[i] == prev:
        seq_val += 1
    else:
        seq.append(seq_val)
        seq_val = 1
        prev = s[i]

last_length = 1
prev = s[-1]
for i, c in enumerate(s[::-1][1:]):
    if c != prev:
        last_length += i
        break
seq += [last_length]

ans = 0
for i in range(n):
    if i % 2 == 0:
        # print(seq[i:i+2*k])
        ans = max(ans, sum(seq[i:i+2*k]))
    else:
        # print(seq[i:i+2*k])
        ans = max(ans, sum(seq[i:i+2*k+1]))

print(ans)