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;
const int mod = 1e9 + 7;
#define ll long long
const int maxn = 1000100;

ll inv[maxn];

void get_inv() {
    for (int i = 1; i < maxn; ++i) {
        if (i == 1) inv[i] = 1;
        else {
            inv[i] = (mod - (mod / i) * inv[mod % i]) % mod;
            if (inv[i] < 0) inv[i] += mod;
        }
    }
}

ll modpow(int n, int r) {
    ll ret = 1;
    ll tmp = (ll)n;
    while (r) {
        if (r & 1) {
            ret = ret * tmp;
        }
        tmp *= tmp;
        tmp %= mod;
        ret %= mod;
        r /= 2;
    }
    return ret;
}


int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    get_inv();
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }
    set<int> s;
    for (auto &x : a) s.insert(x);
    map<int, int> m;
    for (auto &x : s) {
        int p = x;
        for (int i = 2; i * i <= p; ++i) {
            if (p % i == 0) {
                int cnt = 0;
                while (p % i == 0) ++cnt, p /= i;
                m[i] = max(m[i], cnt);
            }
        }

        if (p != 1) m[p] = max(m[p], 1);
    }

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

    ll ans = 0;
    for (auto &x : a) {
        ans = (ans + lcm * inv[x] % mod) % mod;
    }
    cout << ans << endl;
    return 0;
}