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: Wrong Answer)

#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
using namespace std;
//#define int long long
#define d  long double
//#define d long double
#define pii pair<int,int>
#define pb push_back
#define N 200003
#define inf (int)1e18
#define haha ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define MAX 10000003 
using namespace std; 
  
typedef long long ll; 
#define F first 
#define S second
const int mod = 1000000007; 
  
int prime[MAX]; 
unordered_map<int, int> max_map; 
  
// Function to return a^n 
int power(int a, int n) 
{ 
    if (n == 0) 
        return 1; 
    int p = power(a, n / 2) % mod; 
    p = (p * p) % mod; 
    if (n & 1) 
        p = (p * a) % mod; 
    return p; 
} 
  
// Function to find the smallest prime factors 
// of numbers upto MAX 
void sieve() 
{ 
    prime[0] = prime[1] = 1; 
    for (int i = 2; i < MAX; i++) { 
        if (prime[i] == 0) { 
            for (int j =2*i ; j < MAX; j += i) { 
                if (prime[j] == 0) { 
                    prime[j] = i; 
                } 
            } 
            prime[i] = i; 
        } 
    } 
} 
  
// Function to return the LCM modulo M 
void lcmModuloM(const ll* ar, int n) 
{ 
  
    for (int i = 0; i < n; i++) { 
        int num = ar[i]; 
        unordered_map<int, int> temp; 
  
        // Temp stores mapping of prime factor to 
        // its power for the current element 
        while (num > 1) { 
  
            // Factor is the smallest prime factor of num 
            int factor = prime[num]; 
  
            // Increase count of factor in temp 
            temp[factor]++; 
  
            // Reduce num by its prime factor 
            num /= factor; 
        } 
  
        for (auto it : temp) { 
  
            // Store the highest power of every prime 
            // found till now in a new map max_map 
            max_map[it.first] = max(max_map[it.first], it.second); 
        } 
    } 
  
    
} 
signed main()
{
    haha
    int n;
    sieve();
    cin>>n;
    ll a=0;
    ll ar[100001];
  //  int l=1;

    for(int i=0;i<n;i++)
    {
        cin>>ar[i];
      
    }
    lcmModuloM(ar,n);
    //for(auto x:max_map)
    //cout<<x.first<<" "<<x.second<<endl;
    //cout<<l<<endl;
     for (int i = 0; i < n; i++) { 
        int num = ar[i]; 
        unordered_map<int, int> temp; 
  
        // Temp stores mapping of prime factor to 
        // its power for the current element 
        while (num > 1) { 
  
            // Factor is the smallest prime factor of num 
            int factor = prime[num]; 
  
            // Increase count of factor in temp 
            temp[factor]++; 
  
            // Reduce num by its prime factor 
            num /= factor; 
        } 
        ll ans=1;
         for (auto it : max_map) { 
  
            // Store the highest power of every prime 
            // found till now in a new map max_map 
            if(ar[i]%it.first!=0)
            ans*=power(it.first,it.second);
            else
            {
           
            int y=abs(temp[it.first]-it.second);
            ans*=power(it.first,y);
            }
           // ans%=mod;
            
        } 
        ans%=mod;
        a+=ans;
        a%=mod;
     }
   cout<<a;
  
   
    
}