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 Lua (Lua 5.3.5) (judge verdict: Accepted)

local mce, msq = math.ceil, math.sqrt
local function getprimes(x)
  local primes = {}
  local allnums = {}
  for i = 1, x do allnums[i] = true end
  for i = 2, x do
    if allnums[i] then
      table.insert(primes, i)
      local lim = x // i
      for j = 2, lim do
        allnums[j * i] = false
      end
    end
  end
  return primes
end


local function getdivisorparts(x, primes)
  local prime_num = #primes
  local tmp = {}
  local lim = mce(msq(x))
  local primepos = 1
  local dv = primes[primepos]
  while primepos <= prime_num and dv <= lim do
    if x % dv == 0 then
      local t = {}
      t.p = dv
      t.cnt = 1
      x = x // dv
      while x % dv == 0 do
        x = x // dv
        t.cnt = t.cnt + 1
      end
      table.insert(tmp, t)
      lim = mce(msq(x))
    end
    if primepos == prime_num then break end
    primepos = primepos + 1
    dv = primes[primepos]
  end
  if x ~= 1 then
    local t = {}
    t.p, t.cnt = x, 1
    table.insert(tmp, t)
  end
  return tmp
end

local function getdivisorCore(divisorparts)
  local t = {}
  local pat = 1
  local len = #divisorparts
  local allpat = 1
  for i = 1, len do
    allpat = allpat * (1 + divisorparts[i].cnt)
  end
  for t_i_pat = 0, allpat - 1 do
    local div = allpat
    local i_pat = t_i_pat
    local ret = 1
    for i = 1, len do
      div = div // (divisorparts[i].cnt + 1)
      local mul = i_pat // div
      i_pat = i_pat % div
      for j = 1, mul do
        ret = ret * divisorparts[i].p
      end
    end
    table.insert(t, ret)
  end
  table.sort(t)
  return t
end

local primes = getprimes(mce(msq(1000000007)))

local n, k = io.read("*n", "*n")
local a = {}
local sum = 0
for i = 1, n do
  a[i] = io.read("*n")
  sum = sum + a[i]
end

local dvp = getdivisorparts(sum, primes)
local dvs = getdivisorCore(dvp)

for id = #dvs, 1, -1 do
  local dv = dvs[id]
  local plus, minus = {}, {}
  local z = 0
  local c = 0
  for i = 1, n do
    local v = a[i] % dv
    if v <= dv - v then
      table.insert(minus, v)
      z = z - v
    else
      table.insert(plus, dv - v)
      z = z + dv - v
      c = c + dv - v
    end
  end
  if 0 < z then
    table.sort(plus)
    for i = #plus, 1, -1 do
      local v = plus[i]
      z = z - dv
      c = c - v
      if z == 0 then break end
    end
  elseif z < 0 then
    table.sort(minus)
    for i = #minus, 1, -1 do
      local v = minus[i]
      z = z + dv
      c = c + (dv - v)
      if z == 0 then break end
    end
  end
  if c <= k then print(dv) break end
end