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.chomp.split(//)

O = []
E = []
e = 0
o = 0
ix = 0
while ix < S.size
  o = 0
  while ix < S.size && S[ix] == '1'
    o += 1
    ix += 1
  end
  O << o
  break if ix >= S.size

  e = 0
  while ix < S.size && S[ix] == '0'
    e += 1
    ix += 1
  end
  E << e
end
O << 0 if O.size == E.size

if E.size <= K
  v = O.inject(0, &:+) + E.inject(0, &:+)
  puts v
  exit
end

sum = 0
(0..K).each do |i|
  sum += O[i]
end
(0...K).each do |i|
  sum += E[i]
end

max = sum
(K...E.size).each do |i|
  sum -= (O[i - K] + E[i - K])
  sum += (O[i + 1] + E[i])
  max = [max, sum].max
end
p max


# arr = []
# E.size.times do |i|
#   arr[i] = O[i] + E[i] + O[i + 1]
# end

# (K - 1).times do |k|
#   k += 1
#   arr2 = Array.new(E.size, 0)
#   (k...E.size).each do |i|
#     arr2[i] = arr[i - 1] + E[i] + O[i+1]
#   end
#   arr = arr2
# end
# puts arr.max