Problem:
Score : 500 points

Problem Statement

We have a sequence of N integers: A_1, A_2, \cdots, A_N.

You can perform the following operation between 0 and K times (inclusive):

Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.

Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.

Constraints

2 \leq N \leq 500

1 \leq A_i \leq 10^6

0 \leq K \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \cdots A_{N-1} A_{N}

Output

Print the maximum possible positive integer that divides every element of A after the operations.

Sample Input 1

2 3
8 20

Sample Output 1

7

7 will divide every element of A if, for example, we perform the following operation:

Choose i = 2, j = 1. A becomes (7, 21).

We cannot reach the situation where 8 or greater integer divides every element of A.

Sample Input 2

2 10
3 5

Sample Output 2

8

Consider performing the following five operations:

Choose i = 2, j = 1. A becomes (2, 6).

Choose i = 2, j = 1. A becomes (1, 7).

Choose i = 2, j = 1. A becomes (0, 8).

Choose i = 2, j = 1. A becomes (-1, 9).

Choose i = 1, j = 2. A becomes (0, 8).

Then, 0 = 8 \times 0 and 8 = 8 \times 1, so 8 divides every element of A. We cannot reach the situation where 9 or greater integer divides every element of A.

Sample Input 3

4 5
10 1 2 22

Sample Output 3

7

Sample Input 4

8 7
1 7 5 6 8 2 6 5

Sample Output 4

5

Solution in OCaml (4.02.3) (judge verdict: Accepted)

let rec divisors acc acc' i n =
  match compare n (i * i) with
  | -1 -> List.rev_append acc acc'
  | 0 -> List.rev_append acc (i :: acc')
  | 1 ->
      ( if 0 < n mod i
        then divisors acc acc'
        else divisors (i :: acc) (n / i :: acc') ) (i + 1) n
let divisors = divisors [] [] 1

let () = Scanf.scanf "%d %d\n" @@ fun n k ->
  let as_ = Array.init n @@ fun _ -> Scanf.scanf "%d " @@ fun a -> a in
  let ms = Array.make n 0 in
  let acc = Array.make (n + 1) 0 in
  let acc' = Array.make (n + 1) 0 in
  Printf.printf "%d\n" @@
  List.fold_left max min_int @@
  List.filter (fun d ->
    for i = 0 to n - 1 do
      ms.(i) <- as_.(i) mod d;
    done;
    Array.sort compare ms;
    for i = 0 to n - 1 do
      acc.(i + 1) <- acc.(i) + ms.(i)
    done;
    for i = n - 1 downto 0 do
      acc'.(i) <- acc'.(i + 1) + d - ms.(i)
    done;
    Array.fold_left ( || ) false @@
    Array.init (n + 1) @@ fun i ->
      acc.(i) = acc'.(i) && acc.(i) <= k) @@
  divisors @@
  Array.fold_left ( + ) 0 as_