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

#include <bits/stdc++.h>
#include <algorithm>
using namespace std;
typedef long long ll;

ll psum(std::vector<ll> S, int i, int j)
{
  return S[j+1] - S[i];
}



void print(vector<int> v[2])
{
  for(int i=0; i<(int)v[0].size(); i++)
  {
    cout << v[0][i] << " ";
  }
  cout << endl;
  for(int i=0; i<(int)v[1].size(); i++)
  {
    cout << v[1][i] << " ";
  }
  cout << endl;
}

int main()
{
  int n,k;
  cin >> n >> k;
  string S;
  cin >> S;
  std::vector<int> v[2];
  std::vector<ll> ASum;
  int tmp=1,cont=0;

  for(int i=0; i<n; i++)
  {
    int s = S[i]-'0';
    if(s==tmp){
      cont++;
    } else {
      v[tmp].push_back(cont);
      tmp = s;
      cont = 1;
    }
  }
  v[tmp].push_back(cont);
  if(tmp==0) v[1].push_back(0);

  std::vector<ll> psum[2];

  int div = (int)v[0].size();
  ll tmp0=0,tmp1=0;
  for(int i=0; i<div; i++)
  {
    psum[0].push_back(tmp0);
    psum[1].push_back(tmp1);
    tmp0+=v[0][i];
    tmp1+=v[1][i];
  }
  psum[0].push_back(tmp0);
  psum[1].push_back(tmp1);
  tmp1 += v[1][div];
  psum[1].push_back(tmp1);

  ll max=0;
  for(int i=0; i+k<=div; i++)
  {
    ll tmp = psum[0][i+k]-psum[0][i];
    tmp += psum[1][i+k+1]-psum[1][i];
    if(max<tmp) max = tmp;
  }
  if(max==0) cout << n << endl;
  else cout << max << endl;
  return 0;
}