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: Time Limit Exceeded)

N, K = gets.chop.split.map(&:to_i)
S = gets.to_s.chomp
ans = 10000000

# if can generate 11..11 with length can be generate string s with K times operation, return true
#task_possible?("1010101",2, 6) -> false
#task_possible?("1010101",2, 4) -> false
#task_possible?("1010101",2, 5) -> true
def task_possible?(s, k, length)
    ret = false
    slide_window(s,length).each do |fragment|
        if can_transform?(fragment, k) then
            ret = true
        end
    end
    return ret
end


#("10111", 3) -> [101, 011, 111]
#("1011", 2) -> [10, 01, 11]
def slide_window(s, i)
    ret = []
    0.upto(s.length-i) do |j|
        ret.push(s.slice(j..j+i-1))
    end
    return ret
end

# if given string can be changed into 111...1 with k times operation, return True
# (10101, 1) -> False
# (10100, 2) -> True
def can_transform?(s, k)
    cache = {}
    if cache[s] != nil then
        return cache[s]
    end

    if count_zero_lump(s) <= k then
        cache[s] = true
        return true
    else
        cache[s] = false
        return false
    end
end

# count 0 lump
# (00100) -> 2
# (0) -> 1
# (1) -> 0
# (11001) -> 1
def count_zero_lump(s)
    ret = 0
    state = false
    s.chars.each do |c|
        if c=="0" and state==false then
            ret = ret + 1
            state = true
        elsif c=="1" then
            state = false
        end
    end
    return ret
end

y = 0
min = 0
max = 100000
mid = ""
# binary search max length
while true do
    mid = (min + max) / 2
    #p mid
    if task_possible?(S, K, mid) then
        if min==mid then
            break
        end
        min = mid
    else
        if max==mid then
            break
        end
        max = mid
    end
end
p mid