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<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
using namespace std;
typedef long long ll;
typedef unsigned int ui;
const ll mod = 1000000007;
typedef long double ld;
const ll INF = 1e+14;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef complex<ld> Point;
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;
#define fr first
#define sc second
#define all(c) c.begin(),c.end()
#define pb push_back
 
void Yes(){
	cout<<"Yes"<<endl;
	exit(0);
}
 
void No(){
	cout<<"No"<<endl;
	exit(0);
}

int main(){
  ll N,K; cin>>N>>K;
  char S[N];
  rep(i,N){
    cin>>S[i];
  }
  
  vector<ll> hand;
  
  if(S[0]=='0') hand.push_back(0);
  if(N==1) hand.push_back(S[0]-'0');
  else{
    ll count=1;
    rep(i,N-1){
      if(S[i]==S[i+1]){
        count++;
        if(i==N-2) hand.push_back(count);
        continue;
      }
      else{
        hand.push_back(count);
        count=1;
        if(i==N-2) hand.push_back(count);
      }
    }
  }
    if(S[N-1]=='0') hand.push_back(0);
    
    ll M=hand.size();
    ll judge=(M-1)/2;
    
    //累積和
    vector<ll> ad;
    
    ll count1=0;
    ad.push_back(count1);
    rep(i,M){
      count1+=hand[i];
      ad.push_back(count1);
    }
    ll ans=0;
    if(K>=judge) cout<<ad[M]<<endl;
    else{
      ll volu=(M-2*K-1)/2+1;
      rep(i,volu){
        ll volu2=i*2;
        ll ans1=ad[volu2+2*K+1]-ad[volu2];
        ans=max(ans,ans1);
      }
      cout<<ans<<endl;
    }
}