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 <stdio.h>
#include <string>
#include <math.h>
#include <vector>
#include <algorithm>
#include <functional>
#include <map>
#include <climits>
#include <memory>
#include <sstream>
#include <iomanip>
#include <cstdio>

using namespace std;

class Stand {
public:
	int r;
	int num;
};

int main() {

	cin.tie(0);
	ios::sync_with_stdio(false);

	int n, k;
	cin >> n >> k;
	string s;
	cin >> s;
	vector<Stand> stand;
	stand.emplace_back();
	stand.back().r = s[0] == '0' ? 0 : 1;
	stand.back().num = 1;
	for (int i = 1; i < n; i++) {
		if (s[i - 1] == s[i]) {
			stand.back().num++;
		}
		else {
			stand.emplace_back();
			stand.back().r = s[i] == '0' ? 0 : 1;
			stand.back().num = stand[stand.size() - 2].num + 1;
		}
	}

	int max = 0;
	int k_ = s[0] == '0' ? k * 2 - 1 : k * 2;
	if (stand.size() <= k_) {
		cout << n << endl;
		return 0;
	}

	for (int i = 0; i < stand.size() - k_; i++) {
		int min = 0;
		if (i - 1 >= 0)min = stand[i - 1].num;
		max = std::max(max, stand[i + k_].num - min);
	}

	cout << max << endl;

	return 0;
}