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>
//#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
//#include "boost/multiprecision/cpp_int.hpp"
//typedef boost::multiprecision::cpp_int ll;
typedef long double dd;
#define i_7 (ll)(1E9+7)
//#define i_7 998244353
#define i_5 i_7-2

ll mod(ll a){
    ll c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}
typedef pair<ll,ll> l_l;
typedef pair<dd,dd> d_d;
ll inf=(ll)1E16;
#define rep(i,l,r) for(ll i=l;i<=r;i++)
#define pb push_back
ll max(ll a,ll b){if(a<b)return b;else return a;}
ll min(ll a,ll b){if(a>b)return b;else return a;}
void Max(ll &pos,ll val){pos=max(pos,val);}//Max(dp[n],dp[n-1]);
void Min(ll &pos,ll val){pos=min(pos,val);}
void Add(ll &pos,ll val){pos=mod(pos+val);}
dd EPS=1E-9;
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define fi first
#define se second
#define endl "\n"
#define SORT(v) sort(v.begin(),v.end())
#define ERASE(v) v.erase(unique(v.begin(),v.end()),v.end())
#define POSL(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define POSU(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
//template<class T>void max(T a,T b){if(a<b)return b;else return a;}
//template<class T>void min(T a,T b){if(a>b)return b;else return a;}
//template<class T>bool Max(T&a, T b){if(a < b){a = b;return 1;}return 0;}
//template<class T>bool Min(T&a, T b){if(a > b){a = b;return 1;}return 0;}

//////////////////////////



#define MAX_N (ll)2E4///変更！！！！！！！
vector<ll>prime;
bool is_prime[MAX_N+1];
void precal(){
    rep(i,0,MAX_N)is_prime[i]=true;
    is_prime[0]=is_prime[1]=false;
    rep(i,2,MAX_N){
        if(is_prime[i]){
            prime.pb(i);
            for(ll j=2*i;j<=MAX_N;j+=i)is_prime[j]=false;
        }
    }
}
vector<l_l> fact(ll n){
    vector<l_l>v;
    for(ll p:prime){
        if(p*p>n)break;
        ll c=0;
        while(n%p==0){
            n/=p;
            c++;
        }
        if(c)v.pb(l_l(p,c));//pという素因数がc個入っている
    }
    if(n!=1){
        ll vsize=v.size();
        if(vsize>=1&&v[vsize-1].first==n)v[vsize-1].second++;
        else v.pb(l_l(n,1));
    }
    if(v.size()==0){
        v.pb(l_l(1,1));
    }
    return v;
}
/////////////////////////////////////////////////////


ll po(ll i,ll p){
    if(p==0)return 1;
    else{
        i=mod(i);
        if(p==1)return i;
        if(p%2==0)return po(mod(i*i),p/2);
        return mod(i*po(i,p-1));
    }
}
ll bunbo(ll n){
    return po(n,i_5);
}

int main(){fastio
    precal();
    ll n;cin>>n;
    ll a[n];rep(i,0,n-1)cin>>a[i];
    map<ll,ll>mp;
    rep(i,0,n-1){
        vector<l_l>v=fact(a[i]);
        for(auto x:v){
            if(x.fi==1)continue;
            Max(mp[x.fi],x.se);
        }
    }
    ll ret=1;
    for(auto x:mp){
        ret=mod(ret*po(x.fi,x.se));
    }
    //cout<<ret<<endl;
    ll sum=0;
    rep(i,0,n-1){
        Add(sum,ret*bunbo(a[i]));
    }
    cout<<sum<<endl;
    
    return 0;
}