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_arr = gets.chomp.split('').map(&:to_i)

# 最初の色
first_s = s_arr[0]
# 連続した数字のarray
continuity_number_arr = []
# 比較する色
comp_s = s_arr[0]
count_s = 0

s_arr.each do |s|
	if comp_s != s
		continuity_number_arr << count_s
		comp_s = s
		count_s = 1
		next
	end
	count_s += 1
end
continuity_number_arr << count_s

# 最大値
max_count = 0
# arrから抜く数
i = first_s == 1? k*2+1 : k*2
index = 0
continuity_number_arr.each do |num|
	if index % 2 == 1
		index += 1
		next
	end
	last_index = continuity_number_arr.size < index + i ? continuity_number_arr.size - 1 : index + i -1
	nn = continuity_number_arr[index..last_index].inject(:+)
	max_count = nn if nn > max_count
	index += 1
end

puts max_count