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: Runtime Error)

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
 
int main() {
  int n, k;
  string s;
  cin >> n >> k;
  cin >> s;

  vector<int> cluster;
  vector<bool> f;

  char b = s[0];

  cluster.push_back(0);
  f.push_back(false);
  int c_n = 1;
  for(int i=1; i<s.size(); i++) {
    if(b == s[i]) {
      c_n++;
    } else{
      f.push_back(b=='0');
      cluster.push_back(c_n);
      c_n = 1;
      b=s[i];
    }
  }
  cluster.push_back(c_n);
  cluster.push_back(0);
  f.push_back(false);

  int m = f[1] ? cluster[2] : cluster[1];
  for(int i=0; i<cluster.size(); i++) {
    if(f[i]) {
      int c = cluster[i-1] + cluster[i] + cluster[i+1];
      int e = i + 2 * (k-1);
      for(int j=i+2; j<=e+1; j++) {
        c += cluster[j];
      }
      m = c > m ? c : m;
    }
  }

  cout << m << endl;
}