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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;

using E = System.Linq.Enumerable;

public class Program
{
    public static void Main()
    {
        var nk = Console.ReadLine().Split().Select(x => int.Parse(x)).ToArray();
        var n = nk[0];
        var k = nk[1];
        var s = Console.ReadLine().Trim().Select(x => x - '0').ToArray();

        var requirement = new int[s.Length];
        var ls = new int[(s.Length + 1) / 2 + 1];
        var rs = new int[(s.Length + 1) / 2 + 1];
        requirement[0] = s[0] ^ 1;
        for (int i = 1; i < s.Length; i++)
        {
            requirement[i] = requirement[i - 1];
            if (s[i] != s[i - 1])
            {
                if (s[i] == 0)
                {
                    requirement[i]++;
                    rs[requirement[i - 1]] = i;
                }
                else
                {
                    ls[requirement[i]] = i;
                }
            }
        }
        for (int i = requirement.Last(); i < rs.Length; i++)
        {
            rs[i] = s.Length;
        }

        if (k >= rs.Length || rs[k] == s.Length)
        {
            Console.WriteLine(s.Length);
            return;
        }

        for (int i = requirement.Last(); i < ls.Length; i++)
        {
            ls[i] = ls[i - 1];
        }

        var longest = 0;
        for (int i = 0; i < rs.Length; i++)
        {
            if (i + k > rs.Length - 1)
            {
                longest = Math.Max(longest, s.Length - ls[i]);
                break;
            }

            longest = Math.Max(longest, rs[i + k] - ls[i]);
        }
        Console.WriteLine(longest);
    }
}