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: Wrong Answer)

#include <stdio.h>
#include <math.h>

long long gcd(long long a, long long b)
{
	long long temp;

	if (a < b)
	{
		temp = b;
		b = a;
		a = temp;
	}

	while (b != 0)
	{
		temp = b;
		b = a % b;
		a = temp;
	}
	return a;
}

long long gongbeishu(long long a, long long b)
{
	return a * b / gcd(a, b);
}

long long shuzugongbeishu(long long a[], int len)
{
	long long temp[len];
	int i;
	for (i = 0; i < len; i++)
	{
		temp[i] = a[i];
	}

	for (i = 1; i < len; i++)
	{
		temp[i] = gongbeishu(temp[i - 1], temp[i]);
		//printf("%d",temp[i]);
	}
	return temp[len - 1];
}

/*

2, 3, 10

30

*/

int main()
{
	int N, i;
	long long cnt = 0, gongbeishu;
	scanf("%d", &N);
	long long A[N];
	for (i = 0; i < N; i++)
	{
		scanf("%lld", &A[i]);
	}

	gongbeishu = shuzugongbeishu(A, N);

	//printf("%ld\n",gongbeishu);

	for (i = 0; i < N; i++)
	{
		cnt += gongbeishu / A[i];
	}

	printf("%lld\n", cnt % (1000000000 + 7));

	return 0;
}