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)

from collections import deque,defaultdict,Counter
from itertools import accumulate
import bisect
from heapq import heappop,heappush
from fractions import gcd
from copy import deepcopy
import math
import queue
#import numpy as np
Mod = 1000000007




def main(): #startline--------------------------------------------
    N,K = map(int,input().split())
    S = input()
    flag = False
    anslist = []
    if S.count("0") == 0:
        print(N)
        exit()
    if S[0] == "0":
        S = "1"+S
        flag = True
    if S[-1] == "0":
        S += "1" 
        flag = True
    S += "A"
    cnt = 0
    keep = "start"
    for i in S:
        if keep == "start":
            keep = i
        elif keep == i:
            keep = i
        else:
            keep = i
            anslist.append(cnt)
        cnt += 1
    p = len(anslist)
    ans = 0  
    if flag:
        for i in range(p-2*K):
            ans = max(ans,anslist[i+2*K]-anslist[i])
    else:
        for i in range(p-2*K-1):
            ans = max(ans,anslist[i+2*K+1]-anslist[i])
    print(ans)



if __name__ == "__main__":
    main() #endline===============================================





def sieve_of_eratosthenes(n):
    if not isinstance(n,int):
        raise TypeError("n is not int")
    if n<2:
        raise ValueError("n < 2 is not effective")
    prime = [1]*(n+1)
    for i in range(2,int(math.sqrt(n))):
        if prime[i] == 1:
            for j in range(2*i,n):
                if j%i == 0:
                    prime[j] = 0
    res = []
    for i in range(2,n):
        if prime[i] == 1:
            res.append(i)
    return res

class UnionFind:
    def __init__(self,n):
        self.parent = [i for i in range(n+1)]
        self.rank = [0 for i in range(n+1)]
    
    def findroot(self,x):
        if x == self.parent[x]:
            return x
        else:
            y = self.parent[x]
            y = self.findroot(self.parent[x])
            return y
    
    def union(self,x,y):
        px = self.findroot(x)
        py = self.findroot(y)
        if px < py:
            self.parent[py] = px
        else:
            self.parent[px] = py

    def same_group_or_no(self,x,y):
        return self.findroot(x) == self.findroot(y)