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 Julia (0.5.0) (judge verdict: Wrong Answer)

parseInt(x) = parse(Int, x)
parseArray(x::Array{SubString{String}, 1}) = map(parseInt, x)

function main()
  n,k = readline() |> split |> parseArray
  a = readline() |> split |> parseArray

  ans=0
  escape=false
  for d in reverse(divisor(sum(a)))
    d_i=filter(x -> x!=0, sort(a.%d))
    l=length(d_i)
    for i in 1:l
      sum(d_i[1:i])!=sum(d-d_i[i+1:end]) && continue
      sum(d_i[1:i])>k && continue
      ans=d
      escape=true
      break
    end
    escape && break
  end
  println(ans)
end

function divisor(x::Int)
  a=Int[]
  for i in 1:floor(sqrt(x))
    x%i!=0 && continue
    push!(a, i)
    push!(a, x÷i)
  end
  sort!(a)
  a
end

main()