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 Ruby (2.3.3) (judge verdict: Wrong Answer)

n,k=gets.split.map(&:to_i)
a=gets.split.map(&:to_i)
sum=a.inject(0){|s,i| s+=i}
require 'prime'
def divisors(n)
  n.prime_division.inject([1]) do |ary, (p, e)|
    (0..e).map{ |e1| p ** e1 }.product(ary).map{ |a, b| a * b }
  end.sort
end

divs = divisors(sum).reverse

divs.each do |i|
  b=a.map{|j| j%i}.sort
  mae=b.shift
  ushiro=i-b.pop
  until(b.empty?)
    if mae >= ushiro
      ushiro+=b.pop
    else
      mae+=b.shift
    end
  end
  if mae <= k
    puts "#{i} at #{mae} times" 
    exit
  end
end