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 <cstdio>
#include <vector>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <algorithm>
#include <utility>
#include <cstring>
#include <cmath>
using namespace std;

int a[100010];
int d[100010];
int n,k;

bool check(int m){
    for(int i=1;i+m-1<=n;++i){
        if(d[i+m-1]-d[i-1]+(a[i]==0&&a[i-1]==0?1:0)<=k){
            return 1;
        }
    }
    return 0;
}

int main() {
    cin>>n>>k;
    a[0]=1;
    for(int i=1;i<=n;++i){
        scanf("%1d",&a[i]);
        d[i]=d[i-1];
        if(a[i]==0&&a[i-1]==1)d[i]++;
    }
    int lo=0,hi=n,mid;
    int ans=0;
    while(lo<=hi){
        int mid=(lo+hi)/2;
        if(check(mid))ans=mid,lo=mid+1;
        else hi=mid-1;
    }
    cout<<ans<<endl;
    return 0;
}