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>
using namespace std;
#define all(c) (c).begin(),(c).end()
#define pb push_back 
#define mp make_pair
typedef long long ll;

ll modulo=1000000007;

vector<ll> primes;
vector<bool> isp;

ll mul(ll a,ll b, ll c){
	ll ans=0;
	while(b>0){
		if(b&1) ans=(ans+a)%c;
		a=(a+a)%c;
		b>>=1;
	}
	return ans;
}

ll ex2(ll a,ll b,ll c){
	ll ans=1;
	while(b>0){
		if(b&1) ans=mul(ans,a,c);
		a=mul(a,a,c);
		b>>=1;
	}
	return ans;
}

void generate(){
	primes.resize(1e6+10,0);
	isp.resize(1e6+10,1);
	primes[1]=1;
	for(int i=2;i<=1e6;i++){
		if(!isp[i]) continue;
		primes[i]=i;
		for(int j=2;i*j<=1e6;j++){
			isp[i*j]=false;
			if(primes[i*j]==0) primes[i*j]=i;
		}
	}
}



int main(){
	int n; cin>>n;
	generate();
	ll A[n+5]; 
	for(int i=0;i<n;i++) cin>>A[i];
	map<ll,ll> lcm;
	
	for(int i=0;i<n;i++){
		map<ll,ll> div;
		int temp=A[i];
		while(temp!=1){
			ll factor=primes[temp];
			div[factor]++;
			temp/=factor;
		}
		for(auto j:div) lcm[j.first]=max(lcm[j.first],j.second);
	}
	
	ll LCM=1;
	
	for(auto i: lcm)
		LCM=mul(LCM,ex2(i.first,i.second,modulo),modulo);
	
	ll ans=0;
	for(int i=0;i<n;i++){
		ans+=mul(LCM,ex2(A[i],modulo-2,modulo),modulo);
		ans%=modulo;
	}
	cout<<ans;
	
	return 0;
}