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.Collections.Generic;

class Program
{
    static void Main(string[] args)
    {
        var nk = Console.ReadLine().Split();
        var n = int.Parse(nk[0]);
        var k = int.Parse(nk[1]) * 2 + 1;
        var s = Console.ReadLine();

        var d = new List<int> { 1 };
        for (int i = 1; i < s.Length; i++)
        {
            if (s[i - 1] == s[i]) d[d.Count - 1]++;
            else d.Add(d[d.Count - 1] + 1);
        }

        d.Add(d[d.Count - 1]);

        var ans = 0;
        var j = s[0] == '0' ? 1 : 0;

        for (; j < d.Count; j += 2)
        {
            if (k <= j)
            {
                ans = Math.Max(ans, d[j] - d[j - k]);
            }
            else
            {
                ans = Math.Max(ans, d[j]);
            }
        }

        Console.WriteLine(ans);
    }
}