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],0])
else
    if i[n] == i[n-1] then
        next
    else
        m.push([i[n],n])
    end
end
  }
  
m.push([-1,i.count])
kotae =[]
(m.count-1).times{|c|
    (K+1).times{|k|
     if m[c+2*k]==nil then next end
       if m[c][0]==1 and m[c+2*k+1]==nil then next end
    if m[c][0]==0 then
        kotae.push(m[c+2*k][1]-m[c][1])
    else
        kotae.push(m[c+2*k+1][1]-m[c][1])
    end
    }
}
print kotae.max