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)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int ctoi(char in){
  return (int)(in - '0');
}

int main(){
  const ll MOD = 1e9+7;
  ll n;
  cin >> n;
  vector<ll> a(n);
  for(ll i=0; i<n; i++) cin >> a[i];
  
  vector<map<ll,ll>> p(n);
  map<ll, ll> maxs;
  vector<ll> ps;
  for(ll i=0; i<n; i++){
    ll div = a[i];
    bool flag = true;
  	for(ll j=2; j<=sqrt(div+1); j++){
      flag = true;
      while(div % j == 0){
      	p[i][j] += 1;
        div /= j;
      }
      if(a[i] % j == 0){
        flag = false;
        if(maxs.count(j) == 0){
          ps.push_back(j);
          maxs[j] = p[i][j];
        }else{
          maxs[j] = max(maxs[j], p[i][j]);
        }
      }
    }
    if(flag){
      p[i][div] = 1;
      ll j = div;
      if(maxs.count(j) == 0){
        ps.push_back(j);
        maxs[j] = p[i][j];
      }else{
        maxs[j] = max(maxs[j], p[i][j]);
      }
    }
  }
  ll ans = 0;
  for(ll i=0; i<n; i++){
    ll tans = 1;
    for(ll ip : ps){
      tans *= ((ll)pow(ip, maxs[ip]-p[i][ip])%MOD); 
    }
    ans += (tans%MOD);
  }
  cout << ans%MOD << endl;
}