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 <iostream>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <utility>
#include <algorithm>
#include <cstdio>
#include <iomanip>
#include <queue>
#include <deque>
#include <stack>
#include <fstream>
#include <cmath>
#include <random>
#include <complex>
#include <functional>

#define ll int64_t
#define Rep(i, n) for (ll i = 0; i < n; i++)
using namespace std;
typedef vector<ll> vec;
typedef vector<vec> mat;

const ll inf = 1LL << 60;

template<class T> inline void chmin(T& a, T b) {
    if (a > b) {
        a = b;
    }
}
template<class T> inline void chmax(T& a, T b) {
    if (a < b) {
        a = b;
    }
}

vec eratosthenes(ll n) {
    vector<bool> sieve(n+1, true);
    vector<ll> table;
    sieve[0] = sieve[1] = false;
    for (ll i = 2; i <= n; i++) {
        if (sieve[i]) {
            table.push_back(i);
            for (ll j = 2; i * j <= n; j++) {
                sieve[i * j] = false;
            }
        }
    }

    return table;
}

const int MOD = 1e9+7;
class mint {
    ll x;
public:
    mint(ll x = 0) : x((x % MOD + MOD) % MOD) {}
    
    mint operator -() const {
        return mint(-x);
    }
    
    mint &operator +=(const mint &a) {
        if ((x += a.x) >= MOD) x -= MOD;
        return *this;
    }

    mint &operator -=(const mint &a) {
        if ((x += MOD - a.x) >= MOD) x -= MOD;
        return *this;
    }

    mint &operator *=(const mint &a) {
        (x *= a.x) %= MOD;
        return *this;
    }

    mint operator +(const mint &a) const {
        mint res(*this);
        return res += a;
    }

    mint operator-(const mint &a) const {
        mint res(*this);
        return res -= a;
    }

    mint operator *(const mint &a) const {
        mint res(*this);
        return res *= a;
    }

    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const {
        return pow(MOD - 2);
    }

    mint &operator /=(const mint &a) {
        return (*this) *= a.inv();
    }

    mint operator /(const mint &a) const {
        mint res(*this);
        return res /= a;
    }

    friend ostream &operator<<(ostream& os, const mint &m) {
        os << m.x;
        return os;
    }
};


int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll N;
    cin >> N;
    vec A(N);
    Rep (i, N) {
        cin >> A[i];
    }

    vec primes = eratosthenes(1e6+10);

    vector<map<ll, ll>> cnt(N);

    Rep (i, N) {
        for (ll e : primes) {
            ll n = A[i];
            while (n % e == 0) {
                cnt[i][e]++;
                n /= e;
            }
        }
    }

    map<ll, ll> b;

    for (ll e : primes) {
        ll tmp = 0;
        Rep (i, N) {
            chmax(tmp, cnt[i][e]);
        }
        b[e] = tmp;
    }

    mint ans(0);

    Rep (i, N) {
        mint tmp(1);
        for (ll e : primes) {
            tmp *= mint(e).pow(b[e]-cnt[i][e]);
        }
        ans += tmp;
    }

    cout << ans << endl;
}