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: Runtime Error)

n,k = gets.split(" ").map(&:to_i)
s = gets.chomp

a = s.split('0').map(&:length)
b = s.split('1').map(&:length)
a.delete(0)
b.delete(0)

c = []
if b.empty?
  c = a
elsif a.size == b.size
  (a.size-k+1).times do |i|
    t = 0
    (k+1).times {|ii| t+= a[i+ii]}
    c << t
  end

  (a.size-k+1).times do |j|
    t = 0
    (k+1).times {|jj| t+= b[j+jj]}
    c[j] = c[j] + t
  end
elsif a.size > b.size
  (a.size-k).times do |i|
    t = 0
    (k+1).times {|ii| t+= a[i+ii]}
    c << t
  end

  (a.size-k).times do |j|
    t = 0
    k.times {|jj| t+= b[j+jj]}
    c[j] = c[j] + t
  end
else
  (b.size-k+1).times do |i|
    t = 0
    k.times do |ii|
      if a.size == 1
        t+= a[0]
      else
        t+= a[i+ii]
      end
    end
    c << t
  end
  (b.size-k+1).times do |j|
    t = 0
    k.times {|jj| t+= b[j+jj]}
    c[j] = c[j] + t
  end
end

puts c.max