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;

using ll = long long;
static const ll MOD = 1e9 + 7;

// 再帰関数による最大公約数
//template <typename T, typename U>
ll gcd(ll x, ll y) {
  if(y == 0) return x;
  return gcd(y, x % y);
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  int n;
  cin >> n;
  
  vector<ll> a(n);
  for(int i = 0; i < n; i++) cin >> a.at(i);
  
  vector<ll> p(1e6 + 1);
  ll x;
  for(int i = 0; i < n; i++){
    x = a.at(i);
    for(int j = 2; j * j <= a.at(i); j++){
      if(x % j == 0){
        ll keep = 0;
        while(x % j == 0){
          x /= j;
          keep++;
        }
        p.at(j) = max(p.at(j), keep);
      }
    }
    if(x >= 2) p.at(x) = max<ll>(p.at(x), 1);
  }
  
  vector<ll> lcm;
  ll keep = 1;
  for(int i = 2; i <= 1e6; i++){
    if(p.at(i) != 0) keep *= pow(i, p.at(i));
    if(keep > 1e12){
      lcm.push_back(keep);
      keep = 1;
    }
  }
  lcm.push_back(keep);
  
  ll ans = 0;
  for(int i = 0; i < n; i++){
    keep = 1;
    for(int j = 0; j < lcm.size(); j++){
      keep *= lcm.at(j);
      ll gcdkeep = gcd(keep, a.at(i));
      keep /= gcdkeep;
      a.at(i) /= gcdkeep;
      keep %= MOD;
    }
    ans += keep;
    ans %= MOD;
  }
  
  cout << ans << '\n';
}