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;
#define rep(i,n) for (int i = 0; i < (int)(n); ++i)
#define reps(i,n) for (int i = 1; i <= (int)(n); ++i)
#define repd(i,n) for (int i = (int)(n - 1); i >= 0; --i)
#define repds(i,n) for (int i = (int)(n); i > 0; --i)
#define loop(i,x,n) for (int i = (int)(x); i < (n); ++i)
#define loops(i,x,n) for (int i = (int)(x); i <= (n); ++i)
#define loopd(i,x,n) for (int i = (int)(x); i > (n); --i)
#define loopds(i,x,n) for (int i = (int)(x); i >= (n); --i)
#define itrep(i,s) for (auto i = begin(s); i != end(s); ++i)
#define itrepd(i,s) for (auto i = --end(s); i != begin(s); --i)
#define all(f,x,...) [&](decltype(x)& whole) { return (f)(begin(whole), end(whole), ##__VA_ARGS__); }(x)
#define rall(f,x,...) [&](decltype(x)& whole) { return (f)(rbegin(whole), rend(whole), ##__VA_ARGS__); }(x)
using ll = long long;
using ld = long double;
constexpr ll inf = static_cast<ll>(1e17);
constexpr int iinf = static_cast<int>(1e9);
constexpr double dinf = 1e10;
constexpr ld ldinf = 1e17;
ll mod = static_cast<ll>(1e9 + 7);
std::ostream& endn(std::ostream& os) { return os.put(os.widen('\n')); }
template<class T> constexpr int sz(const T& a) { return (int)a.size(); }
template<class T> constexpr T pow2(const T a) { return a * a; }
template<class T, class ... Args> constexpr T mins(T& a, Args... args) { return a = min<T>({ a, args... }); }
template<class T, class... Args> constexpr T maxs(T& a, Args... args) { return a = max<T>({ a,args... }); }
template<typename A, size_t N, typename T>
void fillArray(A(&array)[N], const T& val) { std::fill((T*)array, (T*)(array + N), val); }

template<class T>
struct ModInt {
    T x;
    ModInt(T x = 0) :x((x + mod) % mod) {}
    ModInt& operator+=(const ModInt a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    ModInt& operator-=(const ModInt a) {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    ModInt& operator*=(const ModInt a) {
        (x *= a.x) %= mod;
        return *this;
    }
    ModInt operator+(const ModInt a) const {
        ModInt res(*this);
        return res += a;
    }
    ModInt operator-(const ModInt a) const {
        ModInt res(*this);
        return res -= a;
    }
    ModInt operator*(const ModInt a) const {
        ModInt res(*this);
        return res *= a;
    }
    ModInt pow(T t) const {
        if (!t) return 1;
        ModInt a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }
    ModInt inv() const {
        return pow(mod - 2);
    }
    ModInt& operator/=(const ModInt a) {
        return (*this) *= a.inv();
    }
    ModInt operator/(const ModInt a) const {
        ModInt res(*this);
        return res /= a;
    }
};
using mint = ModInt<long long>;


int n;
ll a[10005];
map<ll, ll> l;
void solve() {
    cin >> n;
    ll ma = 0;
    rep(i, n) cin >> a[i];

    rep(i, n) {
        ll tmp = a[i];
        rep(j, 1005) {
            if (j < 2) continue;
            ll cnt = 0;
            while (tmp % j == 0) {
                ++cnt;
                tmp /= j;
            }
            if (cnt != 0) maxs(l[j], cnt);
        }
        if (tmp != 0) l[tmp] = 1;
    }

    mint ans = 0;
    rep(i, n) {
        mint res = 1;
        for (auto& mp : l) res *= mint(mp.first).pow(mp.second);
        res *= mint(a[i]).inv();
        ans += res;
    }

    cout << ans.x << endn;

}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    solve();
    cout << flush;
    return 0;
}