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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define fast ios_base::sync_with_stdio(false); cin.tie(0);cout.tie(0);
#define filein freopen("input.txt","r",stdin)
#define fileout freopen("output.txt","w",stdout)
using namespace std;
string str;
int  main()
{
    int n,k;
    scanf("%d %d",&n,&k);
    cin>>str;
    char ch ='0';
    int cnt=0;
    vector<pair<ll,char> > v;
    for(int i=0;i<str.size();i++)
    {
        if(str[i]==ch) cnt++;
        else
        {
            if(cnt)v.pb(mp(cnt,ch));
            if(ch=='1') ch='0';
            else ch='1';
            cnt=1;
        }
    }
    v.pb(mp(cnt,ch));
    int dif=2*k;
    ll ans=0;
    for(int i=0;i<v.size();i++)
    {
        ll temp=0;
        char  e=v[i].second;
        int last=0;
        if(e=='0') last=i+dif-1;
        else last=i+dif;
        for(int j=i;j<=last and j<v.size(); j++)
        {
            temp+=v[j].first;
        }
        ans=max(ans,temp);
    }
    printf("%lld\n",ans);
}