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: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i,n) for(ll i = 0; i < (ll)n; i++)
#define FOR(i,a,b) for (ll i=(a);i<(ll)b; i++)
#define ALL(obj) (obj).begin(),(obj).end()
#define INF 1000000000000000
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
using namespace std;
typedef long long ll;
typedef double db;
typedef string str;
typedef pair<ll,ll> p;
const int MOD = 1000000007;

void print(const std::vector<int>& v)
{
  std::for_each(v.begin(), v.end(), [](int x) {
    std::cout << x << " ";
  });
  std::cout << std::endl;
}

int main ()
{
  ll N,K;cin>>N>>K;
  string S; cin>>S;
  vector<int> pos;
  vector<int> num;
  int postmp = 0;
  char tmp = S[0];
  for(int i=1;i<N;i++){
    if(S[i]!=tmp){
      pos.push_back(i-postmp);
      num.push_back(S[i]-'0');
      postmp = i;
      tmp = S[i];
    }
  }
  pos.push_back(N-postmp);
  num.push_back(S[N-1]-'0');
  ll ans = 0;
  //print(pos);
  for (int i = 0; i < pos.size(); i++)
  {
    if(num[i] == 0){
      ll length;
      if (i==0){
          length = 2*K;
      }else{
        length = 2 * K + 1;
      }
        if (i+length>pos.size()){
          ll tmp = accumulate(pos.begin()+i, pos.end(), 0);
          ans = max(ans,tmp);
        }else{
            ll tmp = accumulate(pos.begin()+i, pos.begin()+i+length, 0);
            ans = max(ans,tmp);
        }
    }
    if(num[i] == 1){
      ll length;
      if (i==0){
          length = 2*K + 1;
      }else{
        length = 2 * K + 1;
      }
        if (i+length>pos.size()){
          ll tmp = accumulate(pos.begin()+i, pos.end(), 0);
          ans = max(ans,tmp);

        }else{
            ll tmp = accumulate(pos.begin()+i, pos.begin()+i+length, 0);
            ans = max(ans,tmp);
        }
    }
  }
  cout<<ans<<endl;
  return 0;
}