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)

open Scanf
open Printf

let readInt () = scanf " %d" (fun x -> x)

let divisors x =
  let l = ref [] in
  let r = ref [] in
  let y = ref 1 in
  while !y * !y <= x do
    if x mod !y = 0 then begin
      l := !y :: !l;
      if x <> !y * !y then r := (x / !y) :: !r
    end;
    incr y
  done;
  List.rev !l @ !r

let cost n a d =
  let open Array in
  let a = map (fun x -> x mod d) a in
  fast_sort compare a;
  let tot = fold_left (+) 0 a in
  assert (tot mod d = 0);
  let m = tot / d in
  let def = ref 0 in
  for i=n-m to n-1 do def := !def + d - a.(i) done;
  !def

let solve n k a =
  let tot = Array.fold_left (+) 0 a in
  let divs = divisors tot in
  List.fold_left (fun cur d -> if cost n a d <= k then d else cur) 1 divs

let main () =
  let n = readInt () in
  let k = readInt () in
  let a = Array.init n (fun _ -> readInt ()) in
  printf "%d\n" (solve n k a)

let () = main ()