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;
int main(){
  int n,k,i,state=1,m=1;
  string s;
  cin>>n>>k>>s;
  int a[4*n]={};
  int count=0;
  for(int i=0;i<n;++i){
    if(i==0){
      a[count]=0;
      count++;
    }
    else{
      if(!(s[i-1]==s[i])){
        a[count]=i;
        count++;
      }
    }
  }
    for(int i=count;i<3*n;++i){
      a[i]=n;
    }
  int tem;
  for(int i=0;i<count;++i){
    int b=a[i];
    if((s[b]-'0')==0){
      tem = a[i+2*k] - a[i];
    }
    else{
      tem = a[i+2*k+1] - a[i];
    }
    m=max(m,tem);
  }
  cout<<m;
  return 0;
}