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)

require 'thread'
n,k = gets.split.map(&:to_i)
s = gets.chomp.split('').map(&:to_i)
s.push (-1)
b = []
cur = s[0]
count_0 = 0
count_1 = 0
count_10 = 0
count = 0
sum = 0
max = 0
num_z = 0
q = Queue.new
0.upto(n) do |i|
  if s[i].zero?
    count_0 += 1
  elsif s[i] == 1
    count_1 += 1
  end
  count += 1
  if s[i] != cur
    if s[i].zero? || s[i-1] == 1
      sum += count_1
      count_10 += count_1
      count_1 = 0
      if num_z == k && s[i].zero?
        unless q.empty?
          sum -= q.pop
        end
      else
        num_z += 1
      end
      max = sum if max < sum
    else
      sum += count_0
      count_10 += count_0
      q.push(count_10)
      count_10 = 0
      count_0 = 0
    end
    cur = s[i]
  end
end
puts max