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

s01 = []

before_c = nil
sp = -1
count0 = 0
S.each_char do |c|
  if c != before_c
    sp += 1
    s01[sp] = {type: c, count: 0}
    before_c = c
    count0 += 1 if c == '0'
  end
  s01[sp][:count] += 1
end
s01.unshift({type: '1', count: 0}) if S[0] == '0'
s01 << {type: '1', count: 0} if S[-1] == '0'

k = [K, count0].min

# 累積和用
s01_sum = [s01[0][:count]]
1.upto(s01.size - 1) do |i|
  s01_sum << s01_sum[i-1] + s01[i][:count]
end

l_max = s01.size - k * 2 - 1
if l_max < 0
  puts S.size
  exit
end

# p s01
# p s01_sum

max = 0
0.upto(l_max) do |l|
  next if l.odd?
  r = l + k * 2
  sum = s01_sum[r] - (l > 0 ? s01_sum[l - 1] : 0)
  # p [l, r, sum, max]
  if sum > max
    max = sum
  end
end

puts max