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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 10;
const int M = 1e6 + 10;
const int MOD = 1e9 + 7;
int n;
int a[N],f[M],kz[M],b[N],zs[M];
long long ans = 0;
long long qpow(long long a,int b)
{
	long long ans1 = 1,res = a;
	while(b){
		if(b & 1) ans1 = ans1 * res % MOD;
		res= res * res % MOD;
		b >>= 1;
	}
	return ans1 % MOD;
}
int main()
{
	scanf("%d",&n);
	int maxlen = -1;
	for(int i = 1;i <= n;i++)scanf("%d",&a[i]);
	for(int i = 1;i <= n;i++)b[i] = a[i],maxlen = max(maxlen,b[i]);
	int cnt = 0;
	for(int i = 2;i <= maxlen;i++)
	{
		if(!f[i])
		{
			zs[++cnt] = i;
			int maxtt = 0;
			for(int j = 1;j <= n;j++)
			{
				int tt = 0;
				while(!(b[j] % i))tt++,b[j] /= i;
				maxtt = max(maxtt,tt);
			}
			kz[i] = maxtt;
			for(int k = 2; k <= maxlen / i;k++)f[i * k] = 1;
		}
	}
	for(int i = 1;i <= n;i++)
	{
		long long tans = 1;	
		for(int j = 1;j <= cnt;j++)
		{
			int zz = zs[j];
			int tt = 0;
			while(!(a[i] % zz))tt++,a[i] /= zz;
			int temp = qpow(zz,kz[zz] - tt);
			tans *= temp;
			tans %= MOD;
		}
		ans += tans;
		ans %= MOD;
	}
	printf("%lld",ans);
	return 0;
}