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 <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <cmath>
#include <stdexcept>
#include <map>
using namespace std;
using ll = long long;

vector<ll> primes();

ll gcd(ll a, ll b){
	return (b > 0)? gcd(b, a % b) : a ;
}

ll mpowi(ll mod, ll a,ll b){
    ll res = 1;
    while(b != 0){
        if(b%2 ==1)res=(res*a)%mod;
        a = (a * a)%mod;
        b /= 2;
    }
    return res;
}

ll mdiv(ll mod, ll a, ll b){
    return (a * mpowi(mod,b,mod-2))%mod;
}

void factorize(map<ll,ll>& res,int a){
    for(int i = 2;i < int(sqrt(a)+1);i++){
        while(a%i == 0){
            res[i]++;
            a /= i;
        }
    }
    if(a > 1)res[a]++;
    
}
int main(){
    ll N;cin>>N;
    vector<ll> v;
    map<ll,ll> lcm;
    for(int i = 0;i < N;i++){
        ll a;cin>>a;
        v.push_back(a);
    }
    for(int i = 0;i < N;i++){
        map<ll,ll> fac;
        factorize(fac,v[i]);
        for(auto& x:fac){
            lcm[x.first] = max(lcm[x.first], x.second);
        }
    }
    // for(auto x:lcm){
    //     cout<<x.first<<x.second<<endl;
    // }
    ll res = 0;
    ll MOD = 1'000'000'007;
    ll mlcm = 1;
    for(auto& x:lcm){
        mlcm = (mlcm * mpowi(MOD, x.first, x.second)) % MOD;
    }
    for(ll i = 0;i < N;i++){
        ll a = v[i];
        ll b = mdiv(MOD,mlcm,a);
        
        res = (res + b) % MOD;
    }
    cout<<res<<endl;

}