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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  s += '?';
  n++;
  vector<pair<int, pair<int, int>>> zeros;
  for (int i = 0; i < n; i++) {
    int j = i;
    while (j < n && s[j] == '0') j++;
    if (j > i) {
      zeros.push_back(make_pair(j - i, make_pair(i, j)));
      i = j - 1;
    }
  }
  sort(zeros.rbegin(), zeros.rend());
  for (int i = 0; i < min(k, (int) zeros.size()); i++) {
    int x = zeros[i].second.first, y = zeros[i].second.second;
    for (int j = x; j < y; j++) {
      s[j] = '1';
    }
  }
  int ans = 0;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') cnt++;
    else {
      ans = max(cnt, ans);
      cnt = 0;
    }
  }
  cout << ans << endl;
  return 0;
}