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{|x|x.to_i}
s = gets.chomp.each_char.map {|x| x == '0' ? 0 : 1}

# stand, reverse, stand...
a = []

tmp = 0
i = 0

tmp = 0
while i < n && s[i] == 1
  tmp += 1
  i += 1
end
a << tmp

while i < n
  tmp = 0
  while i < n && s[i] == 0
    tmp += 1
    i += 1
  end
  a << tmp

  tmp = 0
  while i < n && s[i] == 1
    tmp += 1
    i += 1
  end
  a << tmp
end

b = Array.new(a.size + 1)
b[0] = 0

a.size.times do |i|
  b[i+1] = b[i] + a[i]
end

maximum = 0

[k * 2 + 1, b.size - 1].min.step(b.size - 1, 2) do |i|
  tmp = b[i] - b[[i - (2 * k + 1), 0].max]

  maximum = tmp if tmp > maximum
end

p maximum