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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return 1;
    }
    return 0;
}
typedef long long int ll;

#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
#define endl "\n"
const double EPS = 1e-7;
const int INF = 1 << 30;
const ll LLINF = 1LL << 60;
const double PI = acos(-1);
const ll MOD = 1000000007;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

//-------------------------------------

ll GCD(ll a, ll b) {
    if(b == 0)
        return a;
    else
        return GCD(b, a % b);
}

ll LCM(ll a, ll b) { return a * b / GCD(a, b); }

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n;
    cin >> n;
    vector<ll> a(n);
    ll lcm = 1;
    for(int i = 0; i < n; i++) {
        cin >> a[i];
        lcm = LCM(lcm, a[i]);
    }
    ll ans = 0;
    for(ll i = 0; i < n; i++) {
        (ans += lcm / a[i]) %= MOD;
    }
    cout << ans << endl;
}