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: Runtime Error)

#include<iostream>
#include<vector>
#include<cmath>
#include<algorithm>
#include<utility>
#include<map>
using namespace std;
typedef long long ll;
#define MAX 1000000

#if 1
//√MAXでアクセスができるようなvectorに全て突っ込む
map<ll,ll> lcm_all;
ll sieve_check[MAX+1]={0};//i番目が整数iに対応(1~100000)
//エラトステネスの篩を実装する
void es(){
  //0のところが素数、ここでは-1が素数でない
  for(ll i=2;i<=1000;i++){
    if(sieve_check[i]==0){
      lcm_all[i]=0;
      for(ll j=1;j<=floor(MAX/i);j++){
        if(j!=1) sieve_check[j*i]=-1;
      }
    }
  }
}
#endif

//n個の場合のlcmも2個で比べて順に適用していくだけ
signed main(){
  ll inf=pow(10,9)+7;
  ll n;cin >> n;
  vector<ll> a(n);for(ll i=0;i<n;i++){cin >> a[i];}
  es();
  vector< vector<ll> > prime_factors(n,vector<ll>(1001,0));
  for(ll i=0;i<n;i++){
    for(ll j=2;j<=1000;j++){
      while(a[i]%j==0){prime_factors[i][j]++;a[i]/=j;}
    }
    if(a[i]!=1) prime_factors[i][a[i]]=1;
  }
  for(ll i=0;i<n;i++){
    for(ll j=2;j<=1000;j++){
      lcm_all[j]=max(prime_factors[i][j],lcm_all[j]);
    }
  }
  ll cnt=0;
  for(ll i=0;i<n;i++){
    ll cnt_sub=1;
    for(auto& lcm_all_sub : lcm_all){
      cnt_sub*=pow(lcm_all_sub.first,lcm_all_sub.second-prime_factors[i][lcm_all_sub.first]);cnt_sub%=inf;
    }
    cnt+=cnt_sub;
    cnt%=inf;
  }
  cout << cnt << endl;
}