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;

const int64_t mod=1e9+7;
map< int64_t, int64_t > ans;
map< int64_t, int64_t > prime_factor(int64_t n) {
  map< int64_t, int64_t > ret;
  for(int64_t i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n]++;
  return ret;
}
int64_t modpow(int64_t a, int64_t b){
	int64_t res = 1;
	while (b > 0){
		if (b & 1) res = res * a % mod;
		a = a * a % mod;
		b /= 2;
	}
	return res;
}
int main(){
    int n;
    cin >> n;
    int64_t a[n];
    for(int i=0;i<n;i++){
        cin >> a[i];
        for(auto u:prime_factor(a[i])){
            ans[u.first]=max(ans[u.first],u.second);
        }
    }
    int64_t solve=0;
    int64_t tmp; 
    for(int i=0;i<n;i++){
        tmp=1;
        map< int64_t, int64_t > tmp_pri=prime_factor(a[i]);
        for(auto u:ans){
            tmp*=modpow(u.first,max(0L,u.second-tmp_pri[u.first]));
        }
        solve += tmp;
        solve %= mod;
    }    
    cout << solve << endl;
}