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: Wrong Answer)

#include<iostream>
#include<vector>
#include<cmath>
using namespace std;
using ll = long long;
int main(void){
    ll n,k;
    cin >> n >> k;
    string str;
    cin >> str;

    ll cnt = 0;
    ll i = 0;
    if(str[0]=='1') cnt--;
    while(i<n && cnt<k){
        if(str[i]=='1'&&str[i+1]=='0') cnt++;
        i++;
    }
    ll ans = i;
    ll back_i=0;
    str = str + "0";
    while(i<n){
        while(!(str[i]=='1'&&str[i+1]=='0')&&i<n) i++;
        while(!(str[back_i]=='0'&&str[back_i+1]=='1')&&back_i<n) back_i++;
        back_i++;
        i++;
        //cout << "back_i : " << back_i << "  i : " << i << endl;
        ans = max(ans, i-back_i);

    }
    cout << ans << endl;
    return 0;
}