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;

using ll = long long;
using ld = long double;

using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vs = vector<string>;

typedef pair<ll, ll> P;

#define bit(n) (1LL << (n))

//#define int long long

#define all(v) v.begin(), v.end()
#define sortAl(v) sort(all(v))
#define sortAlr(v)          \
  sort(v.begin(), v.end()); \
  reverse(v.begin(), v.end())

#define rep(i, n) for (ll i = 0; i < n; i++)
#define REP(i, n) for (ll i = 1; i < n; i++)

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define FORm(i, m) for (auto i = m.begin(); i != m.end(); i++)

template <class T>
inline void chmax(T &a, T b) {
  a = std::max(a, b);
}
template <class T>
inline void chmin(T &a, T b) {
  a = std::min(a, b);
}

#define mod (ll)(1e9 + 7)
#define INF LLONG_MAX

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);

  ll n, k;
  cin >> n >> k;

  string s;
  cin >> s;

  ll ans = 0;
  ll r = 0;
  ll flip_cnt = 0;
  rep(l, n) {
    for (; r < n; r++) {
      if (s[r] == '0') {
        flip_cnt++;
        for (; r < n; r++) {
          if (s[r] == '1') {
            r--;
            break;
          }
        }
      }
      if (flip_cnt == k) {
        for (; r < n - 1; r++) {
          if (s[r + 1] == '0') {
            break;
          }
        }
        break;
      }
    }

    if (r >= n) {
      r = n - 1;
    }

    chmax(ans, r - l + 1);

    // 端で探索したら終わり
    if (r == n - 1) {
      break;
    }

    flip_cnt--;
    for (; l < n; l++) {
      if (s[l] == '0') {
        break;
      }
    }
    for (; l < n - 1; l++) {
      if (s[l + 1] == '1') {
        break;
      }
    }
  }

  cout << ans << endl;

  return 0;
}