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.split.map(&:to_i)
S = gets.chomp

if S.match(/^0+$/) || S.match(/^1+$/)
  puts S.size
else
  f1 = S[0] == '1'

# for 1
  b1 = S.split(/0+/).delete_if(&:empty?).map(&:size)

# for 0
  b2 = S.split(/1+/).delete_if(&:empty?).map(&:size)

  b3sum = [0]

  b2.size.times do |i|
    if f1
      s = 0
      s += b3sum[i]
      s += b1[i]
      s += b2[i]
      s += b1[i+1] if i + 1 < b1.size
      b3sum << s
    else
      s = 0
      s += b3sum[i]
      s += b1[i - 1] if i > 0
      s += b2[i]
      s += b1[i] if i < b1.size
      b3sum << s
    end
  end

  max = 0
  max_i = 0

  sm = b3sum.size - K - 1
  sm = 0 if sm < 0

  0.upto(sm) do |i|
    r = i + K
    r = b3sum.size - 1 if b3sum.size <= r
    s = b3sum[r] - b3sum[i]

    if max < s
      max = s
      max_i = i
    end
  end

  result = 0

  sm = max_i + K - 1
  sm = b2.size - 1 if b2.size <= sm

  max_i.upto(sm) do |i|
    if f1
      result += b1[i] + b2[i]
      if i == sm
        result += b1[i + 1] if b1.size > i + 1
      end
    else
      result += b1[i - 1] if i > 0
      result += b2[i]
      if i == sm
        result += b1[i] if b1.size > i
      end
    end
  end

  puts result
end