Problem:
Score : 500 points

Problem Statement

Given are N positive integers A_1,...,A_N.

Consider positive integers B_1, ..., B_N that satisfy the following condition.

Condition: For any i, j such that 1 \leq i < j \leq N, A_i B_i = A_j B_j holds.

Find the minimum possible value of B_1 + ... + B_N for such B_1,...,B_N.

Since the answer can be enormous, print the sum modulo (10^9 +7).

Constraints

1 \leq N \leq 10^4

1 \leq A_i \leq 10^6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

Print the minimum possible value of B_1 + ... + B_N for B_1,...,B_N that satisfy the condition, modulo (10^9 +7).

Sample Input 1

3
2 3 4

Sample Output 1

13

Let B_1=6, B_2=4, and B_3=3, and the condition will be satisfied.

Sample Input 2

5
12 12 12 12 12

Sample Output 2

5

We can let all B_i be 1.

Sample Input 3

3
1000000 999999 999998

Sample Output 3

996989508

Print the sum modulo (10^9+7).

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

# 解説AC(できるといいな)
require 'prime'

n = gets.to_i
a = gets.split.map(&:to_i)
lcm_pd = Array.new(1000000, nil)
n_pd = 0
Prime.each(1000000) do |p|
  lcm_pd[n_pd] = [p, 0]
  n_pd += 1
end

for i in 0...n do
  Prime.prime_division(a[i]).each do |e|
#    d = lcm_pd.assoc(e[0])
    d = nil
    for i in 0...n_pd do
      if lcm_pd[i][0] == e[0]
        d = lcm_pd[i]
        break
      elsif lcm_pd[i][0] > e[0]
        break
      end
    end
    if d.nil?
      lcm_pd[n_pd] = e
      n_pd += 1
    else
      d[1] = e[1] if d[1] < e[1]
    end
  end
end
lcm_pd = lcm_pd.slice(0, n_pd)
lcm_a = Prime.int_from_prime_division(lcm_pd)

sum_b = 0
for i in 0...n do
  sum_b = (sum_b + lcm_a / a[i]) % 1000000007
end
puts sum_b