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 <stdio.h>
#include <iostream>
using namespace std;
int main()
{
	int n, k;
	char s[100001];
	cin >> n;
	cin >> k;
	cin >> s;
	int first = 0, last;
	int i;
	int now = 1;
	int range = 0;
	for (i = 0; i < n; i++)
	{
		s[i] -= '0';
	}
	for (i = 0; i < n; i++)
	{
		if (now != s[i])
		{
			if (now == 0)
			{
				range++;
				if (range == k)
				{
					break;
				}
			}
			now = s[i];
		}
	}
	now = 1;
	int max = i;
	if (max == n)
	{
		cout << max << endl;
		return 0;
	}
	for (last = i; last < n; last++)
	{
		if (now != s[last])
		{
			if (now == 1)
			{
				int fnow = 1;
				for (;; first++)
				{
					if (fnow != s[first])
					{
						fnow = s[first];
						if (fnow == 1)
						{
							break;
						}
					}
				}
			}
			else
			{
			}
			now = s[last];
		}
		if (last - first > max)
		{
			max = last - first;
		}

	}
	cout << (max+1) << endl;
	return 0;
}