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)

def gs() gets.chomp end
def gi() gets.to_i end
def gsmi() gets.chomp.split.map(&:to_i) end
 
def desc(ar) ar.sort!{|x,y|y<=>x} end
def min(a,b) a<=b ? a : b end
def max(a,b) a>=b ? a : b end
def sum(ar) ar.inject(:+) end

N,K = gsmi
S = gs

tmp = []

cnt = 0
now = 1
S.each_char do |s|
	if s.to_i==now
		cnt += 1
	else
		tmp << cnt
		now = 1 - now
		cnt = 1
	end
end
tmp << cnt if cnt!=0

tmp << 0 if S[-1] == '0'

ans = 0
left = 0
right = 2*K
if right > tmp.size
	ans = sum tmp 
else
	now = sum tmp[left..right]
	ans = now
end
until (right+2) > tmp.size do
	now -= (tmp[left] + tmp[left+1])
	now += tmp[right+1] + tmp[right+2]
	left += 2
	right += 2
	ans = max(ans,now)
end
puts ans