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;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long ll;
typedef pair<ll,ll> P;
const ll mod=1000000007;
const ll LINF=1LL<<60;
const int INF=1<<30;

map<ll,ll> lcm;

map<ll,ll> prime_factor(ll n){
	map<ll,ll> res;
	for(ll i = 2; i * i  <= n; i++) {
		while(n % i == 0){
			res[i]++;
			n /= i;
		}
		lcm[i] = max(lcm[i],res[i]);
	}
	if (n != 1){
		res[n] = 1;
		lcm[n] = max(lcm[n],res[n]);
	}
	return res;
}

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


int main(){
	ll n;cin>>n;
	vector<ll> a(n);
	ll m = 0;
	for(int i = 0; i < n; i++) {
		cin>>a[i];
		map<ll,ll> x = prime_factor(a[i]);
		m = max(m,a[i]);
	}
	ll l = 1;
	for(ll i = 2; i<= m; i++) {
		if(lcm[i] != 0){
			l *= mod_pow(i,lcm[i]);
			l %= mod;	
		}	
	}
	


	ll ans = 0;
	for(int i = 0; i < n; i++) {
		ll tmp = l;
		tmp *= mod_pow(a[i],mod - 2);
		ans += tmp;
		ans %= mod;
	}
	cout << ans << endl;
	
	return 0;
}