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 Ruby (2.3.3) (judge verdict: Accepted)

class Bookmark
  attr_reader :index
  attr_accessor :head,:tail
  def initialize(index)
    @index = index
    @head = 0
    @tail = 0
  end
end

N, K = gets.split.map(&:to_i)
S = gets.chomp
B = (0 .. N).select{|i| i.zero? || i == N || S[i - 1] != S[i] }
  .map{|i| Bookmark.new(i)}

B.each_cons(2) do |a,b| 
    if S[a.index] == '1'
      d = b.index - a.index
      a.head = d
      b.tail = d
    end
  end

ans = (2 * K >= B.size) ? N : B.each_index.take_while{|i| (i + 2 * K) < B.size }
  .map{|i| B[i + 2 * K].index - B[i].index + B[i + 2 * K].head + B[i].tail }.max

puts ans