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;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
//#define int long long
typedef long long ll;
typedef pair<int,int> P;

int main(){
    int n,k;
    cin >> n >> k;
    string s;
    cin >> s;
    vector<int> vec;
    vec.push_back(0);
    for(int i=1;i<n;i++){
        if((s[i]=='1' && s[i-1]=='0') || (s[i]=='0' && s[i-1]=='1')) vec.push_back(i);
    }
    int ans=0;
    rep(i,vec.size()){
        if(s[vec[i]]=='0'){
            if(i+2*k<vec.size()) ans=max(ans,vec[i+2*k]-vec[i]);
            else ans=max(ans,n-vec[i]);
        }
        else{
            if(i+2*k+1<vec.size()) ans=max(ans,vec[i+2*k+1]-vec[i]);
            else ans=max(ans,n-vec[i]);
        }
    }
    cout << ans << endl;
}