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: Accepted)

# -*- coding: utf-8 -*-

import sys

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7

N, K = MAP()
S = input()

# ランレングス圧縮
# (後の処理がしやすいように、常に1の区間から始まって1の区間で終わる形にする)
A = []
# 番兵
if S[0] == '0':
    A.append(0)
cnt = 1
cur = S[0]
for i, s in enumerate(S[1:], 1):
    if s != cur:
        A.append(cnt)
        cnt = 1
        cur = s
    else:
        cnt += 1
A.append(cnt)

# 例外処理
if len(A) == 1:
    print(N)
    exit()

# 番兵
if S[-1] == '0':
    A.append(0)

M = len(A)
ans = sm = sum(A[:K*2+1])
# 常に1の区間まで取った方が得なので、indexを2ずつ進める
for i in range(K*2+2, M, 2):
    # 前の0の区間と1の区間を増やす
    sm += A[i-1] + A[i]
    # 後ろの0の区間と1の区間を減らす
    if i-K*2-2 >= 0:
        sm -= A[i-K*2-2]
    if i-K*2-1 >= 0:
        sm -= A[i-K*2-1]
    ans = max(ans, sm)
print(ans)