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: Runtime Error)

n, k = gets.split.map(&:to_i)
row = gets.chomp
zero = row.split('1').map{|it| it.size}.filter{|it| it != 0}
one = row.split('0').map{|it| it.size}.filter{|it| it != 0}

one.prepend(0) if row[0] == '0'  # starts with one
zero.append(0) if one.size > zero.size

sums = one.zip(zero).flatten.inject([0]){|acc, it| acc.push(acc[-1] + it)}
range = (k + 1) * 2 - 1
last_index = sums.size - 1

puts (0..last_index).step(2).map{|left|
    right = [left + range, last_index].min
    sums[right] - sums[left]
}.max