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 <bits/stdc++.h>
using namespace std;
int a[1000000],b[1000000];

int main(){
  string s;
  int c=0,x=0,m=0;
  char p[2] = { '1', '0' };
  int n,k;
  
  cin >> n >> k ;
  cin >> s;
  
  for(int i=0; i<n; i++){
    if( s[i] == p[c%2] ) x++;
    else{
      a[c] = x;
      c++;
      x=1;
    }
  }
    
  /*
  for(int i=1; i<s.size(); i++){
    if( s[i] == s[i-1] ) x++;
    else{
      a[c] = x+1;
      c++;
      x=0;
    }
  }*/
  
  a[c] = x;
  
  b[0] = 0;
  for(int i=1; i<=c+1; i++){
    b[i] = a[i-1] + b[i-1];
  }
  
  /*
  for(int i=0; i<=(c+1)/2*2; i++){
    cout << a[i];
  }
  
  cout << endl;
  
  for(int i=0; i<=c+1; i++){
    cout << b[i];
  }
  
  cout << endl;*/
  
  if( (c+1)/2 <= k ) cout << n << endl;
  else{
    for(int i=0; i<=c+2; i+=2){
      m = max( m , b[2*k+1+i]-b[i] );
    }
    cout << m << endl;
  }
  
  return 0;
}