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)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(vec) vec.begin(),vec.end()
typedef long long ll;
typedef pair<ll,ll> l_l;
typedef pair<int,int> i_i;
const ll MOD=1e9+7;
const int INF=1<<30;

map<ll,ll> pfmp; //各素因数の指数を格納
vector<ll> pfv(0); //異なる素因数を昇順に格納

void PF(ll N){
  ll n=N;
  if(n<=0) return;
  ll exp;
  if(n%2==0){
    if(pfmp[2]==0) pfv.push_back(2);
    exp=0;
    while(n%2==0){
      n/=2;
      exp++;
    }
    if(exp>pfmp[2]) pfmp[2]=exp;
  }
  if(n==1) return;
  for(ll i=3;i<=sqrt(N);i+=2){
    if(n%i==0){
      if(pfmp[i]==0) pfv.push_back(i);
      exp=0;
      while(n%i==0){
        n/=i;
        exp++;
      }
      if(exp>pfmp[i]) pfmp[i]=exp;
    }
  }
  if(n!=1){
    if(pfmp[n]==0){
      pfv.push_back(n);
      pfmp[n]=1;
    }
  } 
}

long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

int main(){
  ll N; cin >> N;
  vector<ll> a(N);
  rep(i,N) cin >> a[i];
  rep(i,N){
    PF(a[i]);
  }
  ll LCM=1;
  rep(i,pfv.size()){
    rep(j,pfmp[pfv[i]]){
      LCM*=pfv[i]%MOD;
      LCM%=MOD;
    }
  }
  /*
  cout << "---" << endl;
  for(auto c:pfv){
    cout << c << ":" << pfmp[c] << endl;
  }
  cout << "---" << endl;
  cout << "LCM:" << LCM << endl;
  */
  ll ans=0;
  rep(i,N){
    ans+=LCM*modinv(a[i],MOD)%MOD;
    ans%=MOD;
  }
  /*
  for(auto c:pfv){
    cout << c << " " << pfmp[c] << endl;
  }
  */
  cout << ans << endl;
}