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: Accepted)

#include <iostream>
#include <fstream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cassert>
#include <utility>
#include <iomanip>
#include <chrono>
#include <random>
#include <bitset>

using namespace std;

#define sz(x) (int) x.size()
#define unique(x) x.erase(unique(x.begin(), x.end()), x.end())
#define all(a) a.begin(), a.end()
#define sqr(x) ((x) * (x)) 
#define y1 aksjdaskdjksjfksdjf
#define left kdnvldvoiwejifejg
#define right lkdsjflksdjfdjfk
#define prev asdasfsadjkjsdfjs
#define tm aklsjdasjdasjdkdjjf
#define lcm hddalkwjeidjjhasda
#define random_shuffle asdasdd

const int MAXN = 10500;
const int mod = (int) 1e9 + 7;

int n;
int a[MAXN];
vector <int> p[MAXN];
map <int, int> mp;
long long lcm = 1;

long long binpow(long long x, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) {
      res = (res * x) % mod;
      p--;
    } else {
      x = (x * x) % mod;
      p /= 2;
    }
  }
  return res % mod;
}

long long inv(long long x) {
  return binpow(x, mod - 2) % mod;
}

int main() { 

  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }

  for (int i = 0; i < n; i++) {
    int cur = a[i];
    for (int j = 2; j * j <= cur; j++) {
      if (cur % j == 0) {
        p[i].push_back(j);
        int cnt = 0;
        while (cur % j == 0) {
          cur /= j;
          cnt++;
        }
        mp[j] = max(mp[j], cnt);
      }
    }
    if (cur > 1) {
      p[i].push_back(cur);
      mp[cur] = max(mp[cur], 1);
    }
  }

  for (auto p : mp) {
    int num = p.first, cnt = p.second;
    lcm = lcm * binpow(num, cnt) % mod;
  }

  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long cur = lcm * inv(a[i]) % mod;
    ans = (ans + cur) % mod;
  }

  cout << ans << endl;

  return 0;
}