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: Accepted)

n, k = gets.chomp.split.map(&:to_i)
a = gets.chomp.split.map(&:to_i)

sum = a.reduce(&:+)

divisors = []
(1..(sum**0.5)).each do |i|
  break if i > sum ** 0.5
  if sum % i == 0
    divisors << [i, sum/i]
  end
end

divisors.flatten.sort.reverse_each do |d|
  rem = a.map { |ai| ai%d }.sort
  rem.delete(0)
  n = rem.length
  if n == 0
    puts(d)
    exit
  end
  sum_minus = 0
  sum_plus = d*n - rem.reduce(&:+)
  (n-1).times do |i|
    sum_minus += rem[i]
    sum_plus -= (d - rem[i])
    next if k < [sum_minus, sum_plus].max
    puts(d)
    exit
  end
end