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)

def cin; gets.split.map(&:to_i) end
def cout(*x); puts x.join(" ") end

# exit if $0 != __FILE__

n, k = cin
s = gets.chop

b = nil
ls = []
s.chars.each do |x|
  if x == b then
    ls[ls.length-1] += 1
  else
    ls << 1
    b = x
  end
end

m = ls.max
i = 0
i = 1 if s[0] == "1" && k % 2 == 1
i = 1 if s[0] == "0" && k % 2 == 0
while i < ls.length
  x = 0
  l = [0, i-k].max
  r = [ls.length-1, i+k].min
  for j in l..r
    x += ls[j]
  end
  m = x if x > m
  i += 2
end

puts m