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<functional>
#include<utility>
#include<cmath>
#include<iomanip>
#include<algorithm>
#include<string>
#include<vector>
using namespace std;

int n, k;
string s;



int main() {
	cin >> n>>k>>s;
	int num = 0;
	vector<int> vec(n, 0);
	if (k%3==0)
	{
		num=(k / 3) * 6 + 1;
	}
	else if(k%3==1){
		num = (k / 3) * 6 + 2;
	}
	else {
		num = (k / 3) * 6 + 5;
	}
	int val;
	if (s[0]=='0')
	{
		val = 0;
	}
	else {
		val = 1;
	}
	int temp = s[0];
	int iter = 0;
	vec[iter]++;
	for (int i = 1; i < n; i++)
	{
		if (s[i]==s[i-1])
		{
			vec[iter]++;
		}
		else {
			iter++;
			vec[iter]++;
		}
	}
	vector<int> nvec(n + 1, 0);
	for (int i = 1; i <= iter+1; i++)
	{
		nvec[i] = nvec[i - 1] + vec[i-1];
	}
	int max = 0;
	for (int i = 0; i <= iter+1-num; i++)
	{
		if (num>iter+1)
		{
			if (max<nvec[iter+1])
			{
				max = nvec[iter + 1]-min(vec[0],vec[iter]);
			}
			break;
		}
		if ((k%3!=1&&(i+val)%2&&max<nvec[i+num] - nvec[i])||( k % 3 == 1&&max < nvec[i + num] - nvec[i]))
		{
			max = nvec[i + num]-nvec[i];
		}
	}
	if (k % 3 != 1 && s[n - 1] == '0'&&iter+2-num>=0)
	{
		if (max<nvec[iter+1]-nvec[iter+2-num])
		{
			max = nvec[iter + 1] - nvec[iter + 2 - num];
		}
	}
	if (num > iter + 1)
	{
		if (max < nvec[iter + 1])
		{
			max = nvec[iter + 1];
		}
	}
	cout <<max<< endl;
}