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;
typedef long long ll;

#ifdef DEBUG
#include "inc/debug.hpp"
#else
#define debug(...) 42
#endif

const int mod = 1000000007;
bool is_prime[1000001];

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

ll modpow(ll a, ll b) {
    if (!b) return 1;
    if (b % 2) return (a * modpow(a, b - 1)) % mod;
    ll t = modpow(a, b / 2);
    return (t * t) % mod;
}

int main() {
    int N;
    cin >> N;
    vector<ll> A(N);
    for (int i = 0; i < N; ++i) cin >> A[i];

    map<ll,int> lcm;
    for (int i = 0; i < N; ++i) {
        auto fct = prime_factor(A[i]);
        for (auto &p: fct) {
            lcm[p.first] = max(lcm[p.first], p.second);
        }
    }

    ll L = 1;
    for (auto &p: lcm) {
        L *= modpow(p.first, p.second);
        L %= mod;
    }

    ll res = 0;
    for (int i = 0; i < N; ++i) {
        res += modpow(A[i], mod-2);
        res %= mod;
    }
    res *= L;
    res %= mod;
    cout << res << endl;

    return 0;
}