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: Runtime Error)

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


int main()
{
  int n,k;
  cin>>n>>k;
  string s;
  cin>>s;
  std::vector<int> vec;
  int cnt=1;
  for (int i = 1; i < n; ++i)
  {
    if(s[i-1]!=s[i]){
      vec.emplace_back(cnt);
      cnt=1;
    }
    else cnt++;
  }
  vec.emplace_back(cnt);
  if(vec.size()<=k+2) {
    cout<<n;
    return 0;
  }
  if(s[0]=='0') vec.insert(vec.begin(),0);
  
  std::vector<int> ans;
  int tmp=vec[0];
  int j=1;
  for (int i = 0; i <k ; ++i)
  {
    tmp+=vec[j]+vec[j+1];
    j+=2;
  }
  ans.emplace_back(tmp);
  for (int i = 0; i <vec.size()-2*k-1 ; i+=2)
  {
    tmp+=vec[j]+vec[j+1];
    tmp-=(vec[i]+vec[i+1]);
    ans.emplace_back(tmp);
    j+=2;
  }
  cout<<*max_element(ans.begin(), ans.end());
  
}