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: Runtime Error)

require 'pp'

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

idx_10 = []
S.each_char.with_index do |c,idx|
  idx_10 << idx if c == '1' and S[idx+1] == '0'
end
idx_10 << S.size-1 if S[-1] == '0'

# pp idx_10

idx_01 = []
idx_01 << 0
S.each_char.with_index do |c,idx|
  next if idx == 0
  idx_01 << idx if c == '1' and S[idx-1] == '0'
end

# pp idx_01

offset = (S[0]=='0' ? K-1 : K)
max = 0
idx_01.each_with_index do |b,idx|
  e = (idx_10[idx+offset] or idx_10.last)
  l = e-b+1
  max = l if l > max
end
  
puts max