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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
int prime(long long int n)
{
	if (n < 2)
		return 0;
	if (n == 2)
		return 1;
	if (n % 2 == 0)
		return 0;
	long long int i;
	for (i = 3; i * i <= n; i += 2)
		if (n % i == 0)
			return 0;
	return 1;
}
long long int count(long long int n, long long int p)
{
	long long int res = 0;
	while (n % p == 0)
	{
		res++;
		n /= p;
	}
	return res;
}
long long int modpow(long long int a, long long int n, long long int p)
{
	if (n < 0)
		return 0;
	long long int res = 1;
	while (n > 0)
	{
		if (n % 2 > 0)
			res = res * a % p;
		a = a * a % p;
		n /= 2;
	}
	return res;
}
int main()
{
	long long int n;
	scanf("%lld", &n);
	long long int i, j;
	long long int a[10004];
	for (i = 0; i < n; i++)
		scanf("%lld", &a[i]);
	long long int p = 1000000007;
	long long int c = 0;
	long long int b[100005];
	for (i = 0; i < 1000006; i++)
	{
		if (prime(i) > 0)
		{
			b[c] = i;
			c++;
		}
	}
	long long int max[100005];
	for (i = 0; i < c; i++)
		max[i] = 0;
	long long int r;
	for (i = 0; i < n; i++)
	{
		for (j = 0; j < c; j++)
		{
			r = count(a[i], b[j]);
			if (max[j] < r)
				max[j] = r;
		}
	}
	long long int ab = 1;
	for (i = 0; i < c; i++)
		ab = ab * modpow(b[i], max[i], p) % p;
	long long int ans = 0;
	for (i = 0; i < n; i++)
		ans = (ans + ab * modpow(a[i], p - 2, p) % p) % p;
	printf("%lld\n", ans);
	return 0;
}