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>
#define rep(i, n) for(long long i = 0; i < n; i++)
using namespace std;
typedef long long ll;
const int MAX_N = 10010;
const int MAX_A = 1000010;
const int mod = 1e9 + 7;


ll calc(ll a, ll b, ll p)
{
	if (b == 0) return 1;
	if (b % 2 == 0)
	{
		ll d = calc(a, b / 2, p);
		return (d * d) % p;
	}
	else
		return (a * calc(a, b - 1, p)) % p;
}

vector<int> f(MAX_A, 0);
vector<int> lcm(MAX_A, 0); 
void isPrime(int n)
{
	f[0] = f[1] = -1;
	for(ll i = 2; i <= n; i++)
	{
		if (f[i]) continue;
		f[i] = i;
		for(ll j = i * i; j <= n; j += i)
		{
			if (!f[j]) f[j] = i;
		}
	}
}

void ft(ll n)
{
	vector<int> cnt(n + 1, 0);
	ll x = n;
	while(x != 1)
	{
		cnt[f[x]]++;
		x /= f[x];
	}
	for(ll i = 1; i <= n; i++)
		lcm[i] = max(lcm[i], cnt[i]);
}

signed main(void)
{
	int n;
	cin >> n;
	int a[n];
	ll ans, l;

	isPrime(1000000);
	rep(i, n) 
	{
		cin >> a[i];
		ft(a[i]);
	}
	l = 1;
	for(int i = 1; i < MAX_A; i++)
	{
		rep(j, lcm[i])
		{
			l *= i;
			l %= mod;
		}
	}
	ans = 0;
	rep(i, n)
	{
		ans += l * calc(a[i], mod - 2, mod);
		ans %= mod;
	}
	cout << ans << endl;
	return 0;
}