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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define ALL(v) v.begin(), v.end()
#define V vector
#define P pair
typedef long double ld;
typedef long long ll;
const int MOD =  1000000007;
const int IINF = INT_MAX;
const ll INF = 1LL << 60;


int main() {
    int n, k; cin >> n >> k;
    string s; cin >> s;

    int cnt = 1;
    V<int> one, zero;
    for(int i = 0; i < n; i++){
        if(s[i] == s[i + 1]) cnt++;
        else{
            if(s[i] == '0') zero.push_back(cnt);
            else one.push_back(cnt);
            cnt = 1;
        }
    }

    int ans = 0;
    for(int i = 0; i < (int)one.size(); i++){
        int tmp = 0, cnt = 0;
        if(!one.empty()){
            while(cnt < k){
                tmp += one[i];
                cnt++;
            }
        }
        if(!zero.empty()){
            for(int j = i; j < k; j++){
                tmp += zero[j];
            }
        }
        ans = max(tmp, ans);
    }
    cout << ans << endl;


    return 0;
}