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 <bits/stdc++.h>
#pragma GCC optimize("unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
typedef long long ll;

int main(){
	ios_base::sync_with_stdio(false); 
	cin.tie(NULL);
	ll n,k;
	cin >> n >> k;
	string arr;
	cin >> arr;
	ll count = 1;
	vector <ll> record;
	for (int z=1;z<n;z++){
		if (arr[z-1] != arr[z]){
			if (arr[z-1] == '0'){
				record.push_back(-count);
				count = 0;
			}
			else {
				record.push_back(count);
				count = 0;
			}
		}
		count++;
	}
	if (count != 0) {
		if (arr[n-1] == '0') count = -count;
		record.push_back(count);
	}
	ll answer = 0;
	ll temp = 0;
	ll s = 0;
	ll e = 0;
	for (int z=0;z<record.size();z++){
		if (record[z] > 0){
			answer += record[z];
		}
		else {
			if (temp == k){
				e = z;
				break;
			}
			answer -= record[z];
			temp++;
		}
	}
	temp = answer;
	for (int z=e;z<record.size();z++){
		if (record[z] < 0){
			temp -= record[z];
			for (;s<record.size();s++){
				if (record[s] > 0){
					temp -= record[s];
				} else {
					temp += record[s];
					s++;
					break;
				}
			}
			answer = max(answer,temp);
		} else {
			temp += record[z];
		}
	}
	if (e == 0){
		cout << answer << endl;
	} else {
		cout << max(answer,temp) << endl;
	}
	cin >> n;
}