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)

io = STDIN
n,k=io.gets.split.map(&:to_i)
s=io.gets.chomp
#  n,k=5,1
#  s="00010"
#  n,k=14,2
#  s="11101010110011"

units= s.each_char.chunk_while{|x,y|x==y}.map{|ar|[ar[0],ar.size]}
#start:0
#0101010
#**
# ***
#   ***
#start:1
#1010101
#***
#  ***
#    ***
#end:0
#101010
#***
#  ***
#    **
if units.select{|x,y|x=="0"}.size<=k
  puts n
else
  units.unshift(["1",0]) if units[0][0]=="0"
  units.push(["1",0]) if units[-1][0]=="0"
  #(units.size-1)/2 :0の数
  #((units.size-1)/2-k+1) :try数
  ans=unit_sum=units[0..(2*k)].inject(0){|tot,unit|tot+unit[1]}
  ((units.size-1)/2-k+1).times do |i|
    true
    unit_sum=unit_sum-units[i][1]-units[i+1][1]+units[i+2*k][1]+units[i+2*k+1][1]
    ans=[unit_sum,ans].max
    true
  end
  puts ans
end