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;
using ll = long long;

const int nMax = 1e6;
const int mod = 1e9 + 7;

class Sieve
{
private:
	int n;
	vector<int> f, primes;

public:
	Sieve(int v = 1) : n(v), f(n + 1)
	{
		f[0] = f[1] = -1;
		for (ll i = 2; i <= n; i++) {
			if (f[i]) continue;
			primes.push_back(i);
			f[i] = i;
			for (ll j = i * i; j <= n; j += i) {
				if (!f[j]) f[j] = f[i];
			}
		}
	}

	vector<pair<int, int>> factor(int x)
	{
		vector<pair<int, int>> res;
		for (const auto& p : primes) {
			int y = 0;
			while (x%p == 0) {
				x /= p;
				y++;
			}
			if (y != 0) res.emplace_back(p, y);
		}
		if (x != 1) res.emplace_back(x, 1);
		return res;
	}
};

ll modPow(ll a, ll x, ll mod)
{
	ll res = 1;
	while (0 < x) {
		if (x & 1) res = res * a%mod;
		a = a * a%mod;
		x >>= 1;
	}
	return res;
}

int main()
{

	int n;
	cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n; i++) cin >> a[i];

	Sieve sieve(nMax);
	map<int, int> mp;
	for (int i = 0; i < n; i++) {
		vector<pair<int, int>> f = sieve.factor(a[i]);
		for (const auto& item : f) mp[item.first] = max(mp[item.first], item.second);
	}

	ll lcm = 1;
	for (const auto& item : mp) {
		for (int i = 0; i < item.second; i++) {
			lcm *= item.first % mod;
			lcm %= mod;
		}
	}

	ll ans = 0;
	for (int i = 0; i < n; i++) {
		int b = lcm * modPow(a[i], mod - 2, mod) % mod;
		ans = (ans + b) % mod;
	}
	cout << ans << endl;
}