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)

k = gets.split(" ")[1].to_i
i = gets.split("").map(&:to_i)
m = Array.new
i.count.times{|n|
if n == 0 then
   m.push([i[0],1])
else
    if i[n] == i[n-1] then
        next
    else
        m.push([i[n],n+1])
    end
end
}
if i[-1] == 1 then
m.push([0,i.count+1])
else
  m.push([1,i.count+1])
end
kotae = Array.new
m.count.times{|n|
a = n+2*k
if m[a]== nil then
    next
end
 
  if m[n][0]== 1 and m[a+1]== nil then
    next
  end
 
if m[n][0]== 0 then
    kotae.push(m[a][1] -m[n][1])
else
    kotae.push(m[a+1][1] -m[n][1])
end
 
}
print kotae.max