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 pb push_back
#define mp make_pair
#define mod 1000000007
typedef long long ll;
#define PI 3.14159265
ll exponent(ll x,ll n,ll m)
{  if(n==0)
    return 1;
    if(n%2!=0)
    {
        return (x*exponent(((x%m)*(x%m))%m,(n-1)/2,m))%m;
    }
    else
    return exponent(((x%m)*(x%m))%m,n/2,m);
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(NULL);
    ll n;
    cin>>n;
    ll a[n];
    for(int j=0;j<n;j++)
        cin>>a[j];
    vector<map<ll,ll>>v;
    map<ll,ll>m1;
    for(ll j=0;j<n;j++)
    {
        ll e=a[j];
        map<ll,ll>m;
        for(ll i=2;i*i<=e;i++)
        {
           while(e%i==0)
           {
             ++m[i];
             m1[i]=1;
             e/=i;
           }
        }
        if(e!=1)
        {
            ++m[e];
            m1[e]=1;
        }
        v.pb(m);
    }
    //cout<<"good "<<v.size()<<endl;
    //cout<<s.size()<<endl;
    vector<ll>ans;
    for(auto it:m1)
    {
        if(it.second==1){
        ll maxi=v[0][it.first];
        for(ll i=1;i<n;i++)
        {
            maxi=max(maxi,v[i][it.first]);
        }
        for(ll i=0;i<maxi;i++)
            ans.pb(it.first);
        }
    }
    //cout<<ans.size()<<endl;
    /*for(ll j=0;j<ans.size();j++)
        ans[j]=ans[j]%mod;*/
    ll lcm=1;
    for(ll j=0;j<ans.size();j++)
    {
        lcm=(lcm*ans[j])%mod;

    }
    ll sum=0;
    for(ll j=0;j<n;j++)
    {
        sum+=lcm*exponent(a[j],mod-2,mod);
        sum=sum%mod;
    }
    cout<<sum;
    return 0;
}