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)

def compress(n, s)
  list = []
  
  obj = [s[0].to_i, 0]
  
  n.times do |i|
    if obj[0] != s[i].to_i then
      list.push(obj.dup)
      
      obj[0] = s[i].to_i
      obj[1] = 1
    else
      obj[1] += 1
    end
  end
  
  list.push(obj.dup)
  
  return list
end

def main(argv)
  (n, k) = gets.chomp.split(' ').map(&:to_i)
  s = gets.chomp
  
  list = compress(n, s)
  
  cumsum = [0] * (list.size + 1)
  list.size.times do |i|
    cumsum[i + 1] = cumsum[i] + list[i][1]
  end
  
  if list[0][0] == 0 then
    if list.size < (k + 2) then
      puts cumsum[list.size].to_s
    else
      max = 0
      
      i = 0
      while i < list.size do
        m = -1
        if list[i][0] == 0 then
          m = [list.size, (k + (k + 0)) + i].min
        else # list[i][0] == 1
          m = [list.size, (k + (k + 1)) + i].min
        end
        
        max = [max, cumsum[m] - cumsum[i]].max
        
        i += 1
      end
      
      puts max.to_s
    end
  elsif list[0][0] == 1 then
    if (list.size < (k + 3)) then
      puts cumsum[list.size].to_s
    else
      max = 0
      
      i = 0
      while i < list.size do
        m = -1
        if list[i][0] == 0 then
          m = [list.size, (k + (k + 0)) + i].min
        else # list[i][0] == 1
          m = [list.size, (k + (k + 1)) + i].min
        end
        
        max = [max, cumsum[m] - cumsum[i]].max
        
        i += 1
      end
      
      puts max.to_s
    end
  end
end

if self.to_s == 'main' then
  main(ARGV)
end