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: Wrong Answer)

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

namespace D
{
    class Program
    {
        static void Main(string[] args)
        {
            var NK = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
            var N = NK[0];
            var K = NK[1];
            var S = Console.ReadLine();

            var j = new List<int>()
            {
                0
            };
            var current = S[0];
            foreach (var c in S)
            {
                if (c == current)
                {
                    j[j.Count - 1]++;
                }
                else
                {
                    j.Add(1);
                    current = c;
                }
            }

            var sum = 0;
            var max = 0;
            var count = 0;
            var mcount = 0;
            for (var i = 1; i < j.Count; i += 2)
            {
                count++;

                sum += j[i];

                if (i - 1 >= 0)
                {
                    sum += j[i - 1];
                }

                if (count >= K)
                {
                    var s = sum;
                    if (i + 1 < j.Count)
                    {
                        s += j[i + 1];
                    }
                    max = Math.Max(max, s);
                    mcount += 2;
                    sum -= j[mcount - 1];
                    sum -= j[mcount - 2];
                }
            }

            Console.Write(max);
        }
    }
}