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: Accepted)

#include<bits/stdc++.h>
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define MOD	1000000007
#define int	long long
#define pii pair<int,int> 
#define vi vector<int>
using namespace std;

int modpow(int a, int b, int mod){
	int res = 1;
	a %= mod;
	while(b > 0){
		if(b&1)
			res = (res*a)%mod;
		a = (a*a)%mod;
		b >>= 1;
	}
	return res;
}

int modinv(int a, int m){
	return modpow(a,m-2,m);
}
const int mod = 1e9 + 7;

map<int,int> getDiv(int n){
	map<int,int> m;
	for(int i = 2; i*i <= n; i++){
		if(n % i == 0)
			while(n % i == 0){
				m[i]++;
				n /= i;
			}
	}
	if(n > 1)
		m[n]++;
	return m;
}

int32_t main(){
	int n;
	cin >> n;
	
	int a[n];
	for(int i = 0; i < n; i++)
		cin >> a[i];
	map<int,int> maxi;
	for(int i = 0; i < n; i++){
		map<int,int> m = getDiv(a[i]);
		for(auto x : m){
			if(x.S > maxi[x.F])
				maxi[x.F] = x.S;
		}
	}
	
	int lcm = 1;
	
	for(auto x : maxi){
		lcm = (lcm * modpow(x.F,x.S,MOD)) % MOD;
	}
	int ans = 0;
	for(int i = 0; i < n; i++){
		ans = (ans + lcm * modinv(a[i],MOD)) % MOD;
	}
	
	ans %= mod;
	if(ans < 0)
		ans += MOD;
		
	cout << ans << endl;

	return 0;
}