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
   character(100000)s
   integer n,k
   integer ::i,ma=1,counter=0,j=1
   integer ::a(100000)=0
   read(*,*)n,k
   read(*,*)s
   do i=1,n
      if(mod(j,2)==1)then
        if(s(i:i)=="1")then
          a(j)=a(j)+1
        else
          j=j+1
        endif
        cycle
      else
        if(s(i:i)=="0")then
          a(j)=a(j)+1
        else
          j=j+1
        endif
      endif
      cycle
   enddo
   a(2:j)=a(2:j)+1
   if(j<=2*k)then
      write(*,*)n
   endif
   do i=1,n-2*k,2
      if(ma<sum(a(i:i+k*2)))ma=sum(a(i:i+k*2))
   enddo
   write(*,'(i0)')ma
end program main