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: Wrong Answer)

n, k = gets.chomp.split.map(&:to_i)
s = gets.chomp
one = s.scan(/1{1,}/)
zero = s.scan(/0{1,}/)
if one.size == 0 || zero.size == 0
  p n
  exit
end
len = Array.new(one.size + zero.size, 0)
p one
p zero
ptr = 0
if s[0] == "1"
  (one.size).times do |i|
    len[ptr] = one[i].size
    ptr += 1
    len[ptr] = zero[i].size if zero[i] != nil
    ptr += 1
  end
else
  (zero.size).times do |i|
    len[ptr] = zero[i].size
    ptr += 1
    len[ptr] = one[i].size if one[i] != nil
    ptr += 1
  end
end
p len
o = [k*2, one.size, zero.size].min
n = len.size - o
ans = Array.new(n, 0)
n.times do |i|
  ans[i] += len[i]
  o.times do |j|
    ans[i] += len[i+j+1]
  end
  ans[i] += len[i+o] if s[0] == "1"
end
p ans.max