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: Wrong Answer)

n, k = gets.chomp.split.map(&:to_i)
array = gets.chomp.split('').map(&:to_i)

a = []
last_number = 1
count = 0
n.times do |i|
  if array[i] == last_number
    count += 1
  else
    a << count
    count = 1
    last_number = array[i]
  end

  if i == n - 1
    unless array[i] == last_number
      count = 1
    end

    a << count
  end
end

total =  2 * k + 1
if total >= n
  puts n
  exit
end

n = a.length
answer_array = []
last = 0
i = 0
n.times do
  if i + total - 1 >= n
    break
  end

  if i == 0
    # last = a[0..total-1].sum
    last = 0
    answer_array << last
    i += 2
    next
  end
  tmp = a[i + total - 1] + a[i + total - 2] - a[i - 1] - a[i - 2]
  last += tmp
  answer_array << last
  i += 2
end

puts answer_array.max