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: Time Limit Exceeded)

#include<bits/stdc++.h>

using namespace std;

const long long mod = 1e9 + 7;

long long pow(long long p, long long e){
  if(e == 0) return 1ll;
  else if(e & 1){
    long long r = pow(p, e >> 1);
    return r * r % mod * p % mod;
  }else{
    long long r = pow(p, e >> 1);
    return r * r % mod;
  }
}

int main(){
  int N;
  cin >> N;
  vector<map<long long, long long>> A(N + 1);
  for(int i = 1; i <= N; i++){
    long long a;
    cin >> a;
    for(long long j = 2; j * j <= a; j++){
      while(a % j == 0){
        if(A[i].count(j)){
          A[i][j]++;
        }else{
          A[i][j] = 1;
        }
        a /= j;
      }
    }
    if(a > 1) A[i][a] = 1;
  }
  
  for(int i = 1; i <= N; i++){
    for(auto m : A[i]){
      if(A[0].count(m.first)){
        A[0][m.first] = max(A[0][m.first], A[i][m.first]);
      }else{
        A[0][m.first] = A[i][m.first];
      }
    }
  }
  
  long long sum = 0;
  for(int i = 1; i <= N; i++){
    long long B = 1;
    for(auto m : A[0]){
      if(!A[i].count(m.first)) A[i][m.first] = 0;
      B *= pow(m.first, m.second - A[i][m.first]);
      B %= mod;
    }
    sum += B;
    sum %= mod;
  }
  
  cout << sum << endl;
  
  return 0;
}