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 Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

program main
implicit none
integer :: n, k
character :: s*100000

integer :: i, j, nc, chank, c, m, a
integer , allocatable :: t( : ), u( : )
read(*,*) n, k
read(*,*) s(1:n)
allocate( t(n+1) )

chank = 0
nc = 1
t = 0
do i = 2, n-1
   if( s(i:i) == s(i-1:i-1) ) then
      nc = nc + 1
   else
      chank = chank+1
      t( i ) = nc
      nc = 1 
   end if
end do
if( s(n:n) == s(n-1:n-1) ) then
   chank = chank + 1
   t( n+1 ) = nc+1
else
   t( n ) = nc
   t( n+1 ) = 1
   chank = chank + 2
end if
allocate( u(chank) )

nc = 0
u  = 0
do i = 1, n+1
   if( t( i ) /= 0 ) then
      nc = nc+1
      u(nc) = t(i)
   end if
end do

if( s(1:1) == '1' ) then
   c = 2*k+1
   if( c .ge. chank ) then
      write(*,*) n
      stop
   else
      m = 0
      do j = k+1, chank-k
         if( m .le. sum( u(max(1,j-k):min(chank,j+k)) ) ) then
            m = sum( u(max(1,j-k):min(chank,j+k)) )
         end if
      end do
      write(*,'(i0)') m
   end if
   
else
   c = 2*k
   if( c .ge. chank ) then
      write(*,'(i0)') n
      stop
   else
      m = 0
      do j = k+1, chank-k
         if( max(1,j-k) == 1 ) then
            m = sum(u(1:2*k) )
         else if( m .le. sum( u(max(2,j-k):min(chank,j+k)) ) ) then
            m = sum( u(max(2,j-k):min(chank,j+k)) )
         else if( min( chank, j+k) == chank ) then
            if( m.le. sum( u(chank-2*k+1:chank) ) ) then
               m = sum( u(chank-2*k+1:chank) )
               stop
            end if
         end if
      end do
      write(*,'(i0)') m
   end if
end if
end program main