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;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; i++)


int main () {
  int n, k;
  string s;

  cin >> n >> k >> s;

  int acc = 0;
  vector<int> a, b;
  int i = 0, l = s.size();
  while(i < l) {
    if(i == 0) {
      while(s[i] == '1') {
        acc++;
        i++;
      }
    }

    int c1 = 0;
    int c0 = 0;
    while(i < l && s[i] == '0') {
      c0++;
      i++;
    }
    while(i < l && s[i] == '1')  {
      c1++;
      i++;
    }
    a.push_back(c1 + c0);
    b.push_back(c1);
  }

  vector<int> dpa(k+1, -1);
  vector<int> dpm(k+1, -1);
  dpa[k] = acc;
  dpm[k] = acc;

  for(int i = 0, l = a.size(); i < l; i++) {
    for(int j = 0; j < k; j++){
      if(dpa[j+1] == -1) continue;

      dpa[j] = dpa[j+1] + a[i];
      dpm[j] = max(dpm[j], dpa[j+1] + a[i]);
    }
    dpa[k] = b[i];
    dpm[k] = max(dpm[k], b[i]);
  }

  int ans = 0;
  for(int i = 0; i <= k; i++) ans = max(ans, dpm[i]);

  cout << ans << endl;

  return 0;
}