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: Accepted)

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

int main(int argc, char const *argv[])
{
	int n,k;
	string s;
	cin>>n>>k>>s;
	int f=1,cnt=0;
	if (s.at(0)==0)
	{
		f=0,cnt=1;
	}
	int l=0,r=0;
	while (r<n)
	{
		if (s.at(r)=='0')
		{
			if (f==1)
			{
				f=0;
				++cnt;
				if (cnt>k)
				{
					break;
				}
			}
		}
		else
		{
			if (f==0)
			{
				f=1;
			}
		}
		++r;
	}
	int ans=r-l;
	f=1;
	if (s.at(0)=='0')
	{
		f=0;
	}
	while (r<n)
	{
		while (1)
		{
			if (s.at(l)=='0')
			{
				if (f==1)
				{
					f=0;
				}
			}
			else
			{
				if (f==0)
				{
					break;
				}
			}
			++l;
		}
		while (r<n)
		{
			if (s.at(r)=='0')
			{
				if (f==1)
				{
					break;
				}
			}
			else
			{
				if (f==0)
				{
					f=1;
				}
			}
			++r;
		}
		ans=max(ans,r-l);
	}
	cout<<ans<<endl;
	return 0;
}