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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using static System.Console;
using static System.Math;

namespace E_ABC152
{
    class MainClass
    {

        static int[] s;

        static long[] n;
        public static void Main(string[] args)
        {
            s = Array.ConvertAll(ReadLine().Split().ToArray(), int.Parse);
            n = Array.ConvertAll(ReadLine().Split().ToArray(), long.Parse);
            long tmp = n[0];
            for (int i = 0; i < s[0] - 1; i++)
            {
                tmp = Lcm(tmp, n[i + 1]);
            }

            //WriteLine(tmp);
            long d = 1000000007;
            long ans = 0;
            for (int i = 0; i < s[0]; i++)
            {
                ans = ans +  (tmp / n[i]);
                
                ans = ans % d;
                //WriteLine(ans);
            }
            WriteLine(ans);
        }



        public static long Gcd(long a, long b)
        {
            if (a < b)
                // 引数を入替えて自分を呼び出す
                return Gcd(b, a);
            while (b != 0)
            {
                var remainder = a % b;
                a = b;
                b = remainder;
            }
            return a;
        }

        public static long Lcm(long a, long b)
        {
            return a * b / Gcd(a, b);
        }
    }
}