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;

static class Program{
    static void Main() {
        int[]nk=Console.ReadLine().Split().Select(int.Parse).ToArray();
        int n=nk[0];
        int k=nk[1];
        String s=Console.ReadLine();
        List<int>list=new List<int>();
        if(s[0]!='1')list.Add(0);
        list.Add(0);
        for(int i=1;i<n;i++){
            if(s[i-1]!=s[i])list.Add(i);
        }
        for(int i=0;i<2*k+1;i++)list.Add(n);

        n=list.Count;
        int ans=0;
        for(int i=0;i+2*k+1<n;i+=2){
            ans=Math.Max(ans, list[i+2*k+1]-list[i]);
        }
        Console.WriteLine(ans);
    }
}

/*
0 個の 1 を付け加えて 101010101 型に帰着します。
偶数番目から 2*k+1 個の和です。
rle のようなことをしますが、累積和です。
番兵を 2*k 個入れておきましょう。
*/