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;

#define int long long

const int MOD = 1000000000 + 7;
map<int, int> mp;

void addMaping(int n){

  map<int, int> curr;
  while(n%2 == 0){
    curr[2]++;
    n /= 2;
  }

  for(int i = 3; i * i <= n; i+=2){
    while(n%i == 0){
        curr[i]++;
        n /= i;
    }
  }
  if(n > 2){
    curr[n]++;
  }

  for(auto it : curr){
    if(mp[it.first] < it.second){
        mp[it.first] += abs(it.second - mp[it.first]);
    }
  }
}

signed main(){
  int n; cin >> n;
  vector<int > v(n);

  for(auto &it : v) cin >> it;
  vector<pair<int, int> > a(n);

  int k = 1;

  a[0] = {1, 1};
  vector<int > b;
  b.push_back(1);
  for(int i = 1; i < n; i++){
    int _gcd = __gcd(v[0], v[i]);
    a[i] = {v[0]/_gcd, v[i]/_gcd};
    b.push_back(a[i].second);
  }

  for(int i = 0; i < n; i++){
    addMaping(b[i]);
  }

  //cout << mp.size() << "\n";
  for(auto it : mp){
    //cout << it.first << " " << it.second;
 // cout <<"\n";
    k *= pow(it.first, it.second);
    k %= (MOD);
  }

  int sum = 0;
  for(int i = 0; i < n; i++){
    sum += (a[i].first * (k/a[i].second));
    sum %= MOD;
  }

  cout << sum ;
  return 0;
}