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.Text;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using static System.Console;
using static System.Math;

namespace AtCoder
{
    class Program
    {
        static void Main(string[] args)
        {
            new ABC124().SolveD(new ConsoleInput(Console.In, ' '));
        }
    }

    public class ConsoleInput
    {
        private readonly System.IO.TextReader _stream;
        private char _separator = ' ';
        private Queue<string> inputStream;
        public ConsoleInput(System.IO.TextReader stream, char separator = ' ')
        {
            this._separator = separator;
            this._stream = stream;
            inputStream = new Queue<string>();
        }
        public string Read
        {
            get
            {
                if (inputStream.Count != 0) return inputStream.Dequeue();
                string[] tmp = _stream.ReadLine().Split(_separator);
                for (int i = 0; i < tmp.Length; ++i)
                    inputStream.Enqueue(tmp[i]);
                return inputStream.Dequeue();
            }
        }
        public string ReadLine { get { return _stream.ReadLine(); } }
        public int ReadInt { get { return int.Parse(Read); } }
        public long ReadLong { get { return long.Parse(Read); } }
        public double ReadDouble { get { return double.Parse(Read); } }
        public string[] ReadStrArray(long N) { var ret = new string[N]; for (long i = 0; i < N; ++i) ret[i] = Read; return ret; }
        public int[] ReadIntArray(long N) { var ret = new int[N]; for (long i = 0; i < N; ++i) ret[i] = ReadInt; return ret; }
        public long[] ReadLongArray(long N) { var ret = new long[N]; for (long i = 0; i < N; ++i) ret[i] = ReadLong; return ret; }
    }

    public static class AtCoderLib
    {
        // ユークリッドの互除法 
        public static int Gcd(int a, int b)
        {
            if (a < b)
                // 引数を入替えて自分を呼び出す
                return Gcd(b, a);
            while (b != 0)
            {
                var remainder = a % b;
                a = b;
                b = remainder;
            }
            return a;
        }
    }
  class ABC124
    {
        public void SolveD(ConsoleInput cin)
        {
            var n = cin.ReadInt;
            var k = cin.ReadInt;
            var s = cin.Read;
            var C = s.ToCharArray();
            var A = new List<int>();
            //ランレングス圧縮
            int length = 1;
            for (int i = 1; i < n; i++)
            {
                if (C[i] == C[i - 1])
                {
                    length++;
                }
                else
                {
                    A.Add(length);
                    length = 1;
                }
            }
            A.Add(length);
            //しゃくとり法
            int sum = 0;
            int result = 1;
            int left = 0;
            int right = 2 * k;
            if (C[0] == '1') right++;
            //Write($"A=");
            //foreach (var a in A) { Write(a); }
            //WriteLine();
            sum = A.Skip(left).Take(right - left).Sum();
            while (left < A.Count() - 2 * k)
            {
                //WriteLine($"L={left} R={right} sum={sum}");
                if (sum > result) result = sum;
                sum -= A[left];
                left++;
                if (right - left == 2 * k - 1)
                {
                    right++;
                    if (right <= A.Count()) sum += A[right - 1];
                    right++;
                    if (right <= A.Count()) sum += A[right - 1];
                }
            }
            if (sum > result) result = sum;
            WriteLine(result);
            ReadLine();
        }
    }
}