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>

#define rep(i, n) for (int i = 0; i < int(n); i++)
#define rrep(i, n) for (int i = int(n) - 1; i >= 0; i--)
#define reps(i, n) for (int i = 1; i <= int(n); i++)
#define rreps(i, n) for (int i = int(n); i >= 1; i--)
#define repc(i, n) for (int i = 0; i <= int(n); i++)
#define rrepc(i, n) for (int i = int(n); i >= 0; i--)
#define repi(i, a, b) for (int i = int(a); i < int(b); i++)
#define repic(i, a, b) for (int i = int(a); i <= int(b); i++)
#define each(x, y) for (auto &x : y)
#define all(a) (a).begin(), (a).end()
#define bit(b) (1ll << (b))

using namespace std;

using i32 = int;
using i64 = long long;
using u64 = unsigned long long;
using f80 = long double;
using vi32 = vector<i32>;
using vi64 = vector<i64>;
using vu64 = vector<u64>;
using vf80 = vector<f80>;
using vstr = vector<string>;

inline void yes() { cout << "Yes" << '\n'; exit(0); }
inline void no() { cout << "No" << '\n'; exit(0); }
inline i64 gcd(i64 a, i64 b) { if (min(a, b) == 0) return max(a, b); if (a % b == 0) return b; return gcd(b, a % b); }
inline i64 lcm(i64 a, i64 b) { return a / gcd(a, b) * b; }
inline u64 xorshift() { static u64 x = 88172645463325252ull; x = x ^ (x << 7); return x = x ^ (x >> 9); }
template <typename T> class pqasc : public priority_queue<T, vector<T>, greater<T>> {};
template <typename T> class pqdesc : public priority_queue<T, vector<T>, less<T>> {};
template <typename T> inline void amax(T &x, T y) { if (x < y) x = y; }
template <typename T> inline void amin(T &x, T y) { if (x > y) x = y; }
template <typename T> inline T exp(T x, i64 n, T e = 1) { T r = e; while (n > 0) { if (n & 1) r *= x; x *= x; n >>= 1; } return r; }
template <typename T> inline T bis(T ok, T ng, function<bool(T)> f, T eps = 1) { while (abs(ok - ng) > eps) { T mi = (ok + ng) / 2; (f(mi) ? ok : ng) = mi; } return ok; }
template <typename T> istream& operator>>(istream &is, vector<T> &v) { each(x, v) is >> x; return is; }
template <typename T> ostream& operator<<(ostream &os, vector<T> &v) { rep(i, v.size()) { if (i) os << ' '; os << v[i]; } return os; }
template <typename T, typename S> istream& operator>>(istream &is, pair<T, S> &p) { is >> p.first >> p.second; return is; }
template <typename T, typename S> ostream& operator<<(ostream &os, pair<T, S> &p) { os << p.first << ' ' << p.second; return os; }
void solve(); int main() { ios::sync_with_stdio(0); cin.tie(0); cout << fixed << setprecision(16); solve(); return 0; }

struct Prime {
  int n;
  vector<int> pri;
  Prime() : n(0) {}
  Prime(int n) : n(n) {
    vector<bool> f(n, true);
    f[0] = f[1] = false;
    for (int i = 2; i < n; i++) {
      if (!f[i]) continue;
      pri.push_back(i);
      for (int j = i * 2; j < n; j += i) {
        f[j] = false;
      }
    }
  }
  bool is(int x) {
    if (n > x) {
      return binary_search(pri.begin(), pri.end(), x);
    }
    if (x < 2) return 0;
    for (int i = 2; i * i <= x; i++) {
      if (x % i == 0) return false;
    }
    return true;
  }
  map<int, int> factor(int k) {
    map<int, int> so;
    for (auto p : pri) {
      while (k % p == 0) k /= p, so[p]++;
    }
    if (k != 1) so[k] = 1;
    return so;
  }
};

template <int mod>
struct ModInt {
  int x;
  ModInt(): x(0) {}
  ModInt(long long a) { x = a % mod; if (x < 0) x += mod; }
  ModInt &operator+=(ModInt that) { x = (x + that.x) % mod; return *this; }
  ModInt &operator-=(ModInt that) { x = (x + mod - that.x) % mod; return *this; }
  ModInt &operator*=(ModInt that) { x = (long long) x * that.x % mod; return *this; }
  ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
  ModInt inverse() {
    int a = x, b = mod, u = 1, v = 0;
    while (b) { int t = a / b; a -= t * b; u -= t * v; swap(a, b); swap(u, v); }
    return ModInt(u);
  }
  #define op(o, p) ModInt operator o(ModInt that) { return ModInt(*this) p that; }
    op(+, +=) op(-, -=) op(*, *=) op(/, /=)
  #undef op
  friend ostream& operator<<(ostream &os, ModInt m) { return os << m.x; }
};

using mint = ModInt<1000000007>;

void solve() {
  int n;
  cin >> n;
  vi32 a(n);
  cin >> a;
  auto prime = Prime(1000000);
  vector<unordered_map<int, int>> ve(1000001);
  each(p, prime.pri) {
    for (int i = p; i <= 1000000; i += p) {
      int cnt = 0;
      int ii = i;
      while (ii % p == 0) ii /= p, cnt++;
      ve[i][p] = cnt;
    }
  }
  unordered_map<int, int> tot;
  rep(i, n) {
    each(m, ve[a[i]]) {
      amax(tot[m.first], m.second);
    }
  }
  mint ans = 0;
  rep(i, n) {
    auto ma = tot;
    each(m, ve[a[i]]) {
      ma[m.first] -= m.second;
    }
    mint tmp = 1;
    each(m, ma) {
      tmp *= exp((mint) m.first, m.second);
    }
    ans += tmp;
  }
  cout << ans << endl;
}