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;
using ll=long long;
using ld=long double;

const ll MOD=1000000007;

ll gcd1(ll a,ll b){
  if(a==0 && b==0)return 0;
  ll n;
  while(a%b!=0){
    n=a%b;
    a=b;
    b=n;
  }
  return b;
}

ll solveLinear1(ll n,ll m,ll c){
  if(n==0 && m==0)return 0;
  ll a[100],b[100];
  a[0]=n;b[0]=m;
  int i=1;
  while(b[i-1]!=0){
    a[i]=b[i-1];
    b[i]=a[i-1]%b[i-1];
    i++;
  }
  n=i;
  if(c%a[n-1]!=0){
    return 0;
  }
  ll x=c/a[n-1],y=0;
  for(i=1;i<n;i++){
    ll cnt=x;
    x=y;
    y=cnt-(a[n-i-1]/b[n-i-1])*y;
  }
  return x;
}

ll modRev(ll a,ll p){
  ll n=solveLinear1(a,p,1);
  n=n%p;
  if(n<0)return n+p;
  return n;
}

int main(){
  int i,j,k,cnt=0;
  int n;
  string s;
  cin>>n>>k;
  cin>>s;
  int l[n],r[n];
  if(s[0]=='0'){
    l[cnt]=0;
    r[cnt++]=0;
  }
  for(i=0;i<n;i++){
    if(s[i]=='1'){
      l[cnt]=i;
      while(n-1>i && s[i+1]=='1')i++;
      r[cnt++]=i;
    }
  }
  if(s[n-1]=='0'){
    l[cnt]=n-1;
    r[cnt++]=n-1;
  }
  int m=cnt,ma=0;
  if(m<=k){
    cout<<n<<endl;
    return 0;
  }
  for(i=0;i<m-k;i++){
    ma=max(ma,r[i+k]-l[i]+1);
  }
  cout<<ma<<endl;
}