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++ (Clang 3.8.0) (judge verdict: Wrong Answer)

# include <iostream>
# include <string>
# include <vector>
using namespace std;

int main(){
    int n, k;
    string s;
    cin >> n >>  k >> s;
    
    
    vector<int> vc;
    vc.push_back(0);
    cout << s.size() << endl;
    
    bool state = true;
    try{
        for(int i=0; i<n; i++){
        if(state){
            if(s[i]=='0'){
                vc.push_back(i+1);
                state = false;
            }
        }else{
            if(s[i]=='1'){
                state = true;
            }
        }
    }
    }catch(char *e){
        cout << e << endl;
    }
    
    vc.push_back(n+1);
    int vsize = vc.size();

    // cout << vc.size() << endl;
    // cout << vsize-k << endl;
    // cout << 3281 - 3415 <<endl;

    int max = 0;
    int lsize = vsize - k;
    // cout << lsize << endl;
    
    if(lsize>0){
        for(int i=0; i<lsize; i++){
            // cout << i << endl;
            int fst_index = i;
            int snd_index = i + k + 1;
            if(snd_index>=vc.size()){
                snd_index = vc.size() - 1;
            }
            int tmp = vc[snd_index] - vc[fst_index] - 1;
            if(tmp > max){
                max = tmp;
            }
        }  
    }else{
        max = n;
    }

    
    cout << max << endl;
    return 0;
}