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 = gets.chomp!

x = s.chars.chunk {|ch| ch == "1" }.map {|k, v| [k, v.count] }
x.unshift([false, 0])
x.push([false,0])

num = 0
max = 0

memo = []

loop do
    co = x[num][0] ? k * 2 + 1 : k * 2
    if num + co > x.length
        break
    end

    if memo[num]
        sum = memo[num] + x[num + co - 2][1] + x[num + co - 1][1]
        memo[sum + 2] = sum - x[num][1] - x[num + 1][1]        
    else
        sum = x[num, co].inject(0) {|sum, z| sum + z[1] }
        memo[sum + 2] = sum - x[num][1] - x[num + 1][1]
    end
    max = [max, sum].max
    num += 1
end

puts max