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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
    static long mod = 1000000007;

    public static void main(String[] args) {
        @SuppressWarnings("resource")
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n];
        
        for (int i = 0 ; i < n ; i++) {
            a[i] = Integer.parseInt(sc.next());

        }
        List<Integer> primeList = primeList(1000);
        int[] primeArray = new int[1000007];

        for (int i = 0 ; i < n ; i++) {
            int b = a[i];
            for (int j : primeList) {
                int cnt = 0;
                while (b % j == 0) {
                    cnt++;
                    b /= j;
                }
                primeArray[j] = Math.max(primeArray[j], cnt);
                if (j * j > b) break;
            }
            if (b != 1) {
                primeArray[b] = Math.max(primeArray[b], 1);
                if (!primeList.contains(b)) primeList.add(b);
            }
        }
        long ans = 0;
        for (int i = 0 ; i < n ; i++) {
            long z = 1l;
            int b = a[i];
            for (int p : primeList) {
                int cnt = 0;
                if (primeArray[p] > 0) {
                    while (b % p == 0) {
                        b /= p;
                        cnt++;
                    }
                    z *= pow(p, primeArray[p] - cnt);
                    z %= mod;
                }
            }
            ans += z % mod;
        }
        System.out.println(ans % mod);

    }

    private static List<Integer> primeList(int max) {
        final List<Integer> prime = new ArrayList<Integer>();
        prime.add(2);
        for (int i = 3; i <= max; i += 2) {
            for (int wari : prime) {
                if (i % wari == 0)
                    break;
                if (wari * wari > i) {
                    prime.add(i);
                    break;
                }
            }
        }
        return prime;
    }
    static boolean isPrime(int i) {
        for (int j = 2 ; j <= Math.sqrt(i) ; j++) {
            if (i % j == 0) {
                return false;
            }
        }
        return true;
    }

    static long pow (long x, long n) {
        long sum = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                sum = sum * x % mod;
            }
            x = x * x % mod;
            n >>= 1;
        }
        return sum;
    }

}