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
from itertools import accumulate

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 ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7

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

L=[]
cnt0=0
# 0始まりの時は番兵的に1の区間0を追加しておく
if S[0]=='0':
    L.append(0)
    cur=0
else:
    cur=1
cnt=0
# 1の区間の長さ、0の区間の長さ、1の区間の長さ...の配列Lを作る
for i in range(N):
    if cur==int(S[i]):
        cnt+=1
    else:
        L.append(cnt)
        if cur==0:
            cnt0+=1
        cnt=1
        # フラグの切り替え
        cur=1-cur
else:
    if cnt!=0:
        L.append(cnt)
        if cur==0:
            cnt0+=1
# 0終わりの時は番兵的に1の区間0を追加しておく
if S[-1]=='0':
    L.append(0)

acc=[0]+list(accumulate(L))
mx=0
if cnt0<K:
    # 0区間の数がKより少なかったら、それに合わせて区間和取る長さを短くする
    K=cnt0
for i in range(0, len(acc), 2):
    if i+(K*2+1)<len(acc):
        # 各区間和から最大を取る
        mx=max(mx, acc[i+(K*2+1)]-acc[i])
print(mx)