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: Wrong Answer)

type bin = I of int | O of int

let _ =
  let n, k = Scanf.scanf "%d %d\n" (fun n k -> n, k) in
  let prev = ref (I 0) in
  let res = ref [] in
  for i = 0 to n - 1 do
    let d = Scanf.scanf (if i <> n - 1 then "%1d" else "%1d\n") (fun d -> d) in
    if i = 0 then 
      begin
        prev := if d = 0 then O 1 else I 1;
      end
    else
      begin
        match !prev, d with
        | (I i), 1 -> prev := I (i+1)
        | (O o), 0 -> prev := O (o+1)
        | (I i), 0 -> (prev := O 1; res := (I i)::!res)
        | (O o), 1 -> (prev := I 1; res := (O o)::!res)
        | _ -> failwith "hoge"
      end
  done;
  let ct_of = function I i -> i | O o -> o in
  let input = Array.of_list (!prev::!res) in
  let l = Array.length input in
  let res = ref 0 in
  let i, j, k' = ref 0, ref 0, ref k in
  let temp = ref 0 in
  while !i < l do
(*    Printf.printf "(i:%d, j:%d, k':%d)" !i !j !k';*)
    let d = input.(!j) in
    let j' = match d with | I _ -> min (!j + 2 * !k') (l - 1) | O _ -> min (!j + 2 * !k' - 1) (l - 1) in
    for p = !j to j' do
      temp := !temp + (ct_of input.(p))
    done;
    j := j';
    res := max !res !temp;
    let i' = match input.(!i) with | I _ -> min (!i+2) (l - 1) | O _ -> min (!i+1) (l - 1) in
    k' := 1;
    for p = !i to i' do
      temp := !temp - (ct_of input.(p))
    done;
    i := i';
    if !i = l - 1 then i := l + 10;
  done;
  !res |> print_int