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: Time Limit Exceeded)

N,K = gets.chomp.split(" ").map(&:to_i)
S = gets.chomp.split("").map(&:to_i)

si = 9
counts = []
j = -1

(0..S.count-1).each do |i|
  if si != S[i]
    si = S[i]
    j += 1
    counts[j] = 1
  else
    counts[j] += 1
  end
end

i = 0
j = 0
sum = 0
maxsum = 0

if S[0] == 0
  while j < (K * 2) && j < counts.count
    maxsum += counts[j]
    j += 1
  end
  i += 1
  j = 0
end

while i < counts.count
  while j < (K * 2) + 1 && i+j < counts.count
    sum += counts[i+j]
    j += 1
  end

  if maxsum < sum
    maxsum = sum
  elsif sum == 0
    break
  end

  i += 2
  j = 0
  sum = 0
end

puts maxsum