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 <bits/stdc++.h>

using namespace std;

#define REP(i,a,b) for(int i=a;i<(int)b;i++)
#define rep(i,n) REP(i,0,n)
#define all(c) (c).begin(), (c).end()
#define zero(a) memset(a, 0, sizeof a)
#define minus(a) memset(a, -1, sizeof a)
#define watch(a) { std::cout << #a << " = " << a << "\n"; }
template<class T1, class T2> inline bool minimize(T1 &a, T2 b) { return b < a && (a = b, 1); }
template<class T1, class T2> inline bool maximize(T1 &a, T2 b) { return a < b && (a = b, 1); }
template<class T, class V> istream& operator>> (istream& ist, pair<T, V>& p) { return ist >> p.first >> p.second; }
template<class T> ostream& operator<< (ostream& ost, pair<T, T>& p) { return ost << p.first << ", " << p.second; }
template<class T> istream& operator>> (istream& ist, vector<T>& vs) { for(auto& e: vs) ist >> e; return ist; }

typedef long long ll;
int const inf = INT_MAX / 2;

constexpr ll MOD = 1e9+7;

namespace math { namespace integer {

ll mod_pow(ll x, ll n, ll mod) { if(n == 0) { return 1; } if(n % 2 == 0) { return mod_pow(x * x % mod, n / 2, mod); } else { return x * mod_pow(x, n - 1, mod) % mod; } }
template<class T> T extgcd(T a, T b, T& x, T& y) { T d = a % MOD; if(b != 0) { d = extgcd(b % MOD, a % MOD % b % MOD, y, x); y -= (a / b) * x;} else { x = 1, y = 0; } return d; }
template<class T> T mod_inverse(T x, T mod) { return mod_pow(x, T(mod - 2), mod); /* use fermat */ }
template<class T> T mod_inverse_composite_num_mod(T a, ll mod) { T x, y; extgcd(a, mod, x, y); return (mod + x % mod) % mod; }

}}
using namespace math::integer;

auto prime_factor = [](ll x){map<ll, int> xfact;for(int i=2; (ll)i*i<=x; i++) {while(x % i == 0) {xfact[i]++;x /= i;}}if(x > 1) xfact[x]++;return xfact;};

int main() {
  int N; cin >> N;
  vector<ll> A(N); cin >> A;
  ll x, y;
  map<ll, int> m;
  rep(i, N) {
    auto pf = prime_factor(A[i]);
    for (auto e: pf) {
      m[e.first] = max(m[e.first], e.second);
    }
  }
  ll lcm = 1;
  for (auto e: m) {
    rep(i, e.second) {
      (lcm *= e.first) %= MOD;
    }
  }
  ll ans = 0;
  rep(i, N) {
    ll b = lcm * mod_inverse(A[i], MOD) % MOD;
    (ans += b) %= MOD;
  }
  cout << ans << endl;
}