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: Accepted)

import java.util.Scanner;

public class Main {

	public static final long mod = (long)1e9+7;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] a = new int[n];
		long lcm = 1; long ans = 0; int max = 0;

		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
			max = Math.max(max, a[i]);
		}
		int[] a2 = a.clone();

		int[] fact = new int[max+1];
		for (int i = 0; i < n; i++) {
			for (int j = 2; j*j <= a[i]; j++) {
				int count = 0;
				while(a[i]%j==0) {
					a[i] /= j;
					count++;
				}
				fact[j] = Math.max(fact[j], count);
			}
			if(a[i] != 1) fact[a[i]] = Math.max(fact[a[i]], 1);
		}

		for (int i = 2; i <= max; i++) {
			lcm *= pow(i, fact[i]);
			lcm %= mod;
		}

		for (int i = 0; i < n; i++) {
			long t = lcm*pow(a2[i], mod-2)%mod;
			ans += lcm*pow(a2[i], mod-2);
			ans %= mod;
		}

		System.out.println(ans);
		sc.close();
	}
	public 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;
	}
}