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 C# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;

namespace E
{
    class Program
    {
        static void Main(string[] args)
        {
            Program program = new Program();
            int n = int.Parse(Console.ReadLine());
            string[] instr = Console.ReadLine().Split(' ');
            int[] a = new int[instr.Length];
            long saidaikoyakusuu = 0;
            long sum = 0;

            for(int i = 0; i < n; i++){
                a[i] = int.Parse(instr[i]);
            }

            saidaikoyakusuu = a[0];

            long total = a[0];

            for(int i = 0; i < n; i++){
                saidaikoyakusuu = program.Eucrid(total, a[i]);
                total = total* a[i] / saidaikoyakusuu;
            }

            for(int i = 0; i < n; i++){
                sum += total / a[i];
            }

            sum = sum % (int)(Math.Pow(10, 9) + 7);

            Console.WriteLine(sum);

        }

        long Eucrid(long a, long b){

            while(true){
                if(a > b){
                    a = a % b;
                }else{
                    b = b % a;
                }
                if(a == 0)return b;
                if(b == 0)return a;
            }
        }
    }
}