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.Linq;
using System.Collections.Generic;

class Dmondai
{
    static void Main()
    {
        var nk = Console.ReadLine().Split(' ').Select(x => Int32.Parse(x)).ToArray();
        var N = nk[0];
        var K = nk[1];
        var S = Console.ReadLine();
        var list = new List<int[]>();
        var currentNum = -1;
        var currentNumCount = 0;
        var zeroCount = 0;
        var isFirstZero = false;
        var allCount = 0;
        for (int i=0;i<N;++i)
        {
            if (currentNum != S[i])
            {
                allCount++;
                if (currentNum == -1) isFirstZero = S[i] == '0';
                if (S[i] == '0') zeroCount++;
                if (currentNum != -1) list.Add(new int[2]{i==0?S[i]:currentNum,currentNumCount});
                currentNum = S[i];
                currentNumCount = 1;
            }
            else currentNumCount++;
            if (i == N - 1) list.Add(new int[2] { currentNum, currentNumCount });
        }
        var ans = 0;
        var count = zeroCount - K+1;
        //isFirstZero 0246 else 1357
        //一番初め
        var c = 0;
        if (isFirstZero)
        {
            var iC = 2 * K;
            for (int i=0;i<iC;++i) if (i < allCount) c += list[i][1];
        }
        else
        {
            var iC = 2 * K + 1;
            for (int i = 0; i < iC; ++i) if (i < allCount) c += list[i][1];
        }
        if (ans < c) ans = c;
        //累積和
        for (int i=1;i<count;++i)
        {
            //引く
            if (i == 1 && isFirstZero)
            {
                c -= list[0][1];
            }//0246
            else if(isFirstZero)
            {
                c -= (list[(i - 1) * 2][1] + list[(i - 1) * 2-1][1]);
            }//1357
            else
            {
                c -= (list[(i - 1) * 2][1] + list[(i - 1) * 2+1][1]);
            }
            //足す
            //0246
            if (isFirstZero)
            {
                var addIndex = (i + K - 1) * 2 - 1;
                c += list[addIndex][1];
                if(addIndex+1<allCount)c+= list[addIndex + 1][1];
            }//1357
            else
            {
                var addIndex = (i + K - 1) * 2 + 1;
                c += list[addIndex][1];
                if(addIndex+1<allCount) c += list[addIndex+1][1];
            }
            if (ans < c) ans = c;
        }
        Console.WriteLine(ans);
    }
}