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.chars

def cumulative_sum(s)
  left = 0
  sums = []
  (s + ['-1']).each.with_index do |ch, i|
    if s[left] != ch
      sum = [s[left], sums.empty? ? 0 : sums[-1][2], i]
      sums.push(sum)
      left = i
    end
  end
  sums
end

sums = cumulative_sum(S)
ans = 0
sums.size.times do |i|
  ch, prev_sum, _ = sums[i]
  count =
    if ch == '0'
      left = i > 0 ? sums[i - 1][1] : 0
      right = sums[[i + (K - 1) * 2 + 1, sums.size - 1].min][2]
      right - left
    else
      left = prev_sum
      right = sums[[i + K * 2 - 1 + 1, sums.size - 1].min][2]
      right - left
    end
  ans = count if count > ans
end
puts ans