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: Time Limit Exceeded)

using System;

class Program
{
    static void Main(string[] args)
    {
        var nk = Console.ReadLine().Split();
        var n = int.Parse(nk[0]);
        var k = int.Parse(nk[1]);
        var s = Console.ReadLine().ToCharArray();
        var ans = dfs(s, k);
        Console.WriteLine(ans);
    }

    static int dfs(char[] s, int k)
    {
        if (k == 0) return m2(s);

        var max = m2(s);
        for (int i = 0; i < s.Length; i++)
        {
            for (int j = i + 1; j <= s.Length; j++)
            {
                max = Math.Max(max, dfs(m1(s, i, j), k - 1));
            }
        }

        return max;
    }

    static int m2(char[] s)
    {
        var max = 0;
        var count = 0;
        for (int i = 0; i < s.Length; i++)
        {
            if (s[i] == '1')
            {
                count++;
                max = Math.Max(max, count);
            }
            else
            {
                count = 0;
            }
        }
        return max;
    }

    static char[] m1(char[] x, int l, int r)
    {
        var y = new char[x.Length];
        for (int i = 0; i < x.Length; i++)
        {
            if (l <= i && i < r)
            {
                y[i] = x[i] == '0' ? '1' : '0';
            }
            else
            {
                y[i] = x[i];
            }
        }
        return y;
    }
}