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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using P = pair<int, int>;

#define rep(i, n) for (int i = 0; i < (int)(n); ++i)

const int mod = 1e9 + 7;

int gcd(int a, int b) { return b ? a % b : a; }

ll modpow(ll k, ll n) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) res = res * k % mod;
    k = k * k % mod;
    n /= 2;
  }
  return res;
}

struct Sieve {
  vector<int> p;
  Sieve(int n) : p(n) {
    for (int i = 2; i <= n; ++i) {
      p[1] = -1;
      if (p[i] == 0) {
        p[i] = i;
        for (int j = 2 * i; j <= n; j += i) {
          if (p[j]) continue;
          p[j] = i;
        }
      }
    }
  }

  map<int, int> factMap(int n) {
    map<int, int> mp;
    while (p[n] != -1) {
      ++mp[p[n]];
      n /= p[n];
    }
    return mp;
  }
};

int main() {
  int n;
  cin >> n;
  int a[n];
  rep(i, n) cin >> a[i];

  Sieve s(1000005);

  map<int, int> mp;
  rep(i, n) {
    auto m = s.factMap(a[i]);
    for (auto p : m) {
      mp[p.first] = max(mp[p.first], p.second);
    }
  }

  ll lcd = 1;
  for (auto p : mp) {
    rep(i, p.second) { lcd = lcd * p.first % mod; }
  }

  int b[n];
  rep(i, n) { b[i] = lcd * modpow(a[i], mod - 2) % mod; }

  ll ans = 0;
  rep(i, n) { ans = (ans + b[i]) % mod; }

  cout << ans << endl;

  return 0;
}