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 <algorithm>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <list>
#include <map>
#include <memory>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <utility>
#include <vector>
using namespace std;

typedef unsigned long ul;
typedef long long ll;
typedef pair<ul, ul> P;
#define REP(i, n) for (int i = 0; i < n; i++)
const int INF = 1 << 29;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N, K;
  string S;
  cin >> N >> K;
  cin >> S;
  vector<int> diff_num;
  diff_num.reserve(N);
  int number = 0;
  diff_num.emplace_back(0);
  deque<int> state_part;
  auto old_num = S[0];
  for (int ii = 1; ii < N; ii++) {
    if (old_num != S[ii]) {
      diff_num.emplace_back(ii);
      old_num = S[ii];
    }
  }
  int diff_point_num = diff_num.size();
  diff_num.emplace_back(N);
  vector<int> max_num(diff_point_num, 0);
  for (int ii = 0; ii < diff_point_num; ++ii) {
    if (S[diff_num[ii]] == '0') {
      max_num[ii] = diff_num[min(diff_point_num, 2 * K + ii)] - diff_num[ii];
    } else {
      max_num[ii] =
          diff_num[min(diff_point_num, 2 * K + 1 + ii)] - diff_num[ii];
    }
  }
  auto ans = *max_element(max_num.begin(), max_num.end());
  cout << ans << endl;

  return 0;
}