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>
using namespace std;

#define int long long
#define rep(i,l,r) for(int i=(int)(l);i<(int)(r);i++)
#define all(x) (x).begin(),(x).end()
#define pb push_back
template<class T>bool chmax(T &a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T>bool chmin(T &a,T b){if(a>b){a=b;return 1;}return 0;}

typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;

const int inf = 1LL<<60;
const int mod = 1e9 + 7;
const double eps = 1e-9;

/*{
}*/

signed main()
{
  int n, k;
  string s;
  cin >> n >> k >> s;

  int cnt = 0, t = 1;
  vi a;
  rep(i, 0, n){
    if(s[i]-'0' == t%2){
      cnt++;
    }else{
      a.pb(cnt);
      t++;
      cnt = 1;
    }
  }
  a.pb(cnt);
  if(a.size()%2 == 0) a.pb(0);
  int m = a.size();

  multiset<int> ms;
  int sum = 0;
  int K = 2*k+1;
  rep(i, 0, K){
    ms.insert(a[i]);
    sum += a[i];
  }

  int ans = sum;
  for(int i = K; i < m; i += 2){
    ms.erase(ms.find(a[i-K]));
    sum -= a[i-K];
    ms.erase(ms.find(a[i-K+1]));
    sum -= a[i-K+1];

    ms.insert(a[i]);
    sum += a[i];
    ms.insert(a[i+1]);
    sum += a[i+1];
    chmax(ans, sum);
//    cout << i << " : " << ans << endl;
  }

  cout << ans << endl;

//  rep(i, 0, m) cout << a[i] << endl;

  return 0;
}