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 OCaml (4.02.3) (judge verdict: Accepted)

open Scanf
open Printf

let (n,k) = sscanf (read_line ()) "%d %d" (fun x y -> (x,y));;
let str = sscanf (read_line ()) "%s" (fun x -> x);;

let string2cnt_array s =
  let len = String.length s in
  let array = Array.make len (0,'0') in
  let rec loop i c num cnt =
    if (i = len) then
      let () = array.(num) <- (cnt,c) in
      let ret = Array.make (num+1) (0,'0') in
      let () = Array.blit array 0 ret 0 (num+1) in
      ret
    else if (s.[i] = c) then
      loop (i+1) c num (cnt+1)
    else
      let () = array.(num) <- (cnt,c) in
      loop (i+1) (s.[i]) (num+1) 1
    in loop 0 s.[0] 0 0;;


let cnt_array = string2cnt_array str;;
let len = Array.length cnt_array;;
let ruiseki = Array.make len 0;;
let () = ruiseki.(0) <- fst cnt_array.(0) in
for i = 1 to len - 1 do
  ruiseki.(i) <- ruiseki.(i-1) + (fst cnt_array.(i))
done

let rec solve i ans =
  if i = len then ans
  else
    let num = if snd cnt_array.(i) = '0' then 0 else 1 in
    let target = 2 * k - (1 - num) + i in
    let si = min (len - 1) target in
    let ruiseki_minus = if i = 0 then 0 else ruiseki.(i-1) in
    let next_ans = max ans (ruiseki.(si) - ruiseki_minus) in
    solve (i+1) next_ans;;

printf ("%d\n") (solve 0 0);;