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 <bits/stdc++.h>

using namespace std;
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define rep(i,b) FOR(i,0,b)
#define INF 1e18
#define dump(x) cerr<<#x<<"="<<x<<endl
#define all(a) (a).begin(),(a).end()
typedef pair<int,int> P;
template <class T> void chmin(T & a, T const & b) { if (b < a) a = b; }
template <class T> void chmax(T & a, T const & b) { if (b > a) a = b; }
using ll = long long;

int n,k;
string s;
vector<int> vec;
int rui[112345];
int ans;
int main(){
    cin>>n>>k;
    cin>>s;
    //連続する0の長さ
    rep(i,(int)s.size()){
        if(s[i]=='0'){
            int ret=1;
            for(int j=i+1;j<(int)s.size();j++,i++){
                if(s[j]=='1')break;
                else ret++;
            }
            vec.push_back(ret);
        }
        if(s[i]=='1'){
            int ret=1;
            for(int j=i+1;j<(int)s.size();j++,i++){
                if(s[j]=='0')break;
                else ret++;
            }
            vec.push_back(ret);
        }
    }

    rep(i,vec.size()){
     //   cout<<vec[i]<<endl;
        rui[i+1]=vec[i];
    }
    rep(i,vec.size()){
        rui[i+1]+=rui[i];
    }

    rep(i,vec.size()+1){
   //     cout<<rui[i]<<endl;
    }



    //s[0]==0 なら vec[i%2==0] の s[i]== 0
    //s[0]==1 なら vec[i%2==0] の s[i]==1

    for(int i=0;i<vec.size()-2*k;i++){
        if(s[0]=='0' ){
            if(i%2==0)ans=max(ans,rui[i+2*k]-rui[i]);
            else ans=max(ans,rui[i+2*k+1]-rui[i]);

        }
        if(s[0]=='1' ){
            if(i%2==0)ans=max(ans,rui[i+2*k+1]-rui[i]);
            else ans=max(ans,rui[i+2*k]-rui[i]);
        }
    }
    cout<<ans<<endl;




  



}