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)

N, K = gets.split.map(&:to_i)
S = gets

cur_n = "1";

num = 0

arr = []

max = 0

prev = false

S.length.times do |i|
  
  if S[i] != cur_n
    arr.push(num)
    num = 1
    cur_n = S[i]
  else
    num += 1
  end
end  


arr.length.times do |i|
  p_ = prev
  
  if(i % 2 == 0)
    if !prev 
      p_ = 0
      (K*2 + 1).times do |j|
        break if i + j >= arr.length
        p_ += arr[i + j].to_i
  
      end
    else
      
      if i + K * 2 - 1 < arr.length

        p_ += arr[i + K * 2 - 1]

        if i + K * 2 < arr.length
          p_ += arr[i + K * 2]
        end
      end
      
      p_ -= (arr[i - 2] + arr[i - 1])
    
    end

  max = p_ if p_ > max
  prev = p_

  else
    next  
  end

end

puts max