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.split.map {|x| x.to_i}
s = gets

r = 2*k+1

c = s[0]
i = 1
cnt = 1
a = []
while i < n
  if c != s[i]
    a << cnt
    c = s[i]
    cnt = 1
  else
    cnt += 1
  end
  i += 1
end
a << cnt

m = a.size
if s[0] == "0"
  m -= 1
end

#p a
if r > m
  sum = 0
  m.times {|i|
    sum += a[i]
  }
  maxv = sum
else

sum = 0
r.times {|i|
  sum += a[i]
}

maxv = sum
if m > r
(m-r).times { |j|
  sum -= a[j]
  sum += a[j+r]
  if sum > maxv
    maxv = sum
  end
}
end
end
p maxv