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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

int main(){
  int N,K;
  string S;

  cin>>N>>K;
  cin>>S;

  vector<int> a;

  int i,count,now;

  now=1;
  count=0;
  for(i=0;i<N;i++){
    if(S.at(i)==(char)('0'+now))
    count++;
    else{
      a.push_back(count);
      now=1-now;
      count=1;
    }
  }
  if(count!=0)
  a.push_back(count);

  int maxsum=0;
  int left=0;
  int right=left;


  for(left=0;left<a.size();left+=2){
    int sum=0;
    int res=0;

    // cout<<"aaa"<<endl;
    while(res<=2*K-1){
      right+=2;
      res=right-left;
    }
    if(right>=a.size()){
      for(i=left;i<a.size();i++)
      sum+=a.at(i);
      if(maxsum<sum)
      maxsum=sum;
      break;
    }
    for(i=left;i<=right;i++)
    sum+=a.at(i);

    if(maxsum<sum)
    maxsum=sum;

    // cout<<maxsum<<endl;
  }

  cout<<maxsum<<endl;


}