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;
typedef long long ll;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;
template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

const long double EPS = 1e-10;
const long long INF = 1e18;
const long double PI = acos(-1.0L);
const ll mod = 1000000007;
map<ll, ll> mp;
ll N;
ll ans;
ll A[10500];

ll beki(ll a, ll b) {
    if(b == 0) return 1;
    ll ret = beki(a, b / 2);
    ret = ret * ret % mod;
    if(b & 1) ret = ret * a % mod;
    return ret;
}

int main() {
    //cout.precision(10);
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> N;
    for(int i = 0; i < N; i++) cin >> A[i];
    for(int i = 0; i < N; i++) {
        ll copy = A[i];
        for(ll j = 2; j * j <= copy; j++) {
            if(copy % j != 0) continue;
            ll num = 0;
            while(copy % j == 0) {
                copy /= j;
                num++;
            }
            chmax(mp[j], num);
        }
        if(copy != 1) {
            chmax(mp[copy], 1LL);
        }
    }
    for(int i = 0; i < N; i++) {
        ll now = 1;
        for(auto tmp : mp) {
            ll num = 0;
            while(A[i] % tmp.first == 0) {
                num++;
                A[i] /= tmp.first;
            }
            //cerr << i << " " << tmp.first << " " << num << endl;
            now *= beki(tmp.first, tmp.second - num);
            now %= mod;
        }
        ans += now;
        ans %= mod;
    }
    cout << ans;
    return 0;
}