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.*;

public class Main {
	static int P = (int)1e9+7;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		boolean[] prime = new boolean[1000000];
		Arrays.fill(prime, true);
		prime[0] = false;
		prime[1] = false;
		for(int i=2;i*i<prime.length;i++) {
			if(prime[i]) {
				for(int j=i*i;j<prime.length;j+=i) {
					prime[j] = false;
				}
			}
		}
		
		ArrayList<Integer> list = new ArrayList<Integer>();
		for(int i=2;i<prime.length;i++) {
			if(prime[i])
				list.add(i);
		}
		
		int n = sc.nextInt();
		int[] a = new int[n];
		int[] cnt = new int[1000000];
		for(int i=0;i<n;i++) {
			int num = sc.nextInt();
			a[i] = num;
			
			int idx = 0, c = 0, d = list.get(idx);
			while(num>1) {
				if(num%d==0) {
					num/=d;
					c++;
				}
				else {
					cnt[d] = Math.max(cnt[d], c);
					idx++;
					d = list.get(idx);
					c = 0;
				}
			}
			cnt[d] = Math.max(cnt[d], c);
		}
		
		long lcm = 1;
		for(int i=2;i<cnt.length;i++) {
			for(int j=0;j<cnt[i];j++) {
				lcm = (lcm*i) % P;
			}
		}

		long ans = 0;
		for(int i=0;i<n;i++) {
			long ainv = fp(a[i], P-2);
			ans = (ans + ainv) % P;
		}
		
		ans = ans * lcm % P;
		
		System.out.println(ans);
	}
	static long fp(int a, int p) {
		if(p==0)
			return 1;
		long num = fp(a, p/2);
		if(p % 2 == 0)
			return num * num % P;
		else
			return num * num % P * a % P;
	}
}