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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <math.h>
#include <iomanip>
using namespace std;


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

  int d[100100];
  d[0] = 0;
  int cnt = 0;
  for(int i = 1; i < s.size(); i++) {
    if(s[i] != s[i-1]) {
      cnt++;
      d[cnt] = i;
    }
  }

  if (cnt == 0) {
    cout << s.size() << endl;
    return 0;
  }

  cnt++;
  d[cnt] = s.size();
  d[cnt+1] = s.size();
  d[cnt+2] = s.size();

  int ans[100100];
  int fans = 0;
  if(s[0] == '0') {
    ans[0] = d[2*k];
    fans = ans[0];
  }
  for (int i = 0; i <= cnt - 2*k + 1; i++) {
    if (s[d[i]] == '1') {
    ans[i] = d[i+2*k+1] - d[i];
    }
    fans = max(fans, ans[i]);
  }

  cout << fans << endl;
}