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 Julia (0.5.0) (judge verdict: Accepted)

parseInt(x)=parse(Int,x)
parseMap(x::Array{SubString{String},1})=map(parseInt,x)

function main()
	n,k = readline() |> split |> parseMap
	s = chomp(readline())
	a = Int[]
	b = Int[0]
	f = s[1]=='0'?0:1
	t = 1
	for i in 2:n
		if s[i-1] != s[i]
			push!(a,f)
			f = f==0?1:0
			push!(b,t)
			b[length(b)] += b[length(b)-1]
			t = 1
		else
			t += 1
		end
	end
	if t != 0
		push!(a,f)
		push!(b,t)
		b[length(b)] += b[length(b)-1]
	end
	l = length(a)
	if l == 1
		println(n)
	else
		m = 0
		if k%2 == 0
			for i in 1:l
				if a[i] == 1
					x = b[min(i+k+1,l+1)]-b[max(i-k,1)]
					m = max(m,x)
				end
			end
		else
			for i in 1:l
				if a[i] == 0
					x = b[min(i+k+1,l+1)]-b[max(i-k,1)]
					m = max(m,x)
				end
			end
		end
		println(m)
	end
end
main()