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: Wrong Answer)

function calc_seq(xs)
    seq = Int[]
    count = 1
    for i = 1:length(xs)-1
        if xs[i] == xs[i+1]
            count += 1
        else
            push!(seq, count)
            count = 1
        end
    end
    push!(seq, count)
    seq
end

function main(n, k, xs)
    seq = calc_seq(xs)
    m = length(seq)
    if xs[1] == 0
        seq = vcat(0, seq)
        m = length(seq)
    end
    if m % 2 == 0
        seq = vcat(seq, 0)
        m = length(seq)
    end
    val = sum(seq[1:min(2k+1, m)])
    biggest = val
    i = 1
    while i <= (m - (2k+1) + 2) ÷ 2
        i += 2
        val += sum(seq[2k+i-1:2k+i]) - sum(seq[i-2:i-1])
        if biggest < val
            biggest = val
        end
    end
    print(biggest)
end

n, k = [parse(Int, x) for x in split(chomp(readline()))]
str = chomp(readline())
xs = zeros(Int, n)

for (i, s) in enumerate(str)
    xs[i] = parse(Int, s)
end

main(n, k, xs)