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 Python (3.4.3) (judge verdict: Runtime Error)

#include <iostream>
#include <algorithm>
#include <climits>
#include <numeric>
#include <iomanip>
#include <cmath>
#include <vector>
#include <set>
#include <unordered_set>
#include <string>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <unordered_map>
 
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define cina(s, n, a) for(int i = s; i < n; i++) cin >> a[i]
#define couta(n, a) for(int i = 0; i < n; i++) cout << a[i] << ' '
#define copy(s, n, a, b) for(int i = s; i < n; i++) b[i] = a[i]
#define yesno(ok) cout << (ok ? "Yes" : "No")
#define mod(a, b) (b + (a % b)) % b
#define MOD  (ll)1e9 + 7
 
typedef long long ll;
typedef unsigned long long ull;
typedef std::vector<int> vi;
typedef std::pair <int, int> pii;
typedef std::pair <ll, ll> pll;
typedef std::vector<pii> vpii;
using namespace std;

ll gcd (ll a, ll b) {
    while (b) {
        a %= b;
        swap (a, b);
    }
    return a;
}
ll lcm (ll arr[], int n)
{
    ll ans = arr[0];
    for (int i = 1; i < n; i++){
        ans = (((arr[i] * ans)) /
                (gcd(arr[i], ans)));
    }
    return ans;
}
 
void solve(){
    int n;
    cin >> n;
    ll a[200500];
    cina(0, n, a);
    ll lc = lcm(a, n);
    
    ll ans = 0;
    for(int i = 0; i < n; i++){
        ans += lc / a[i];
        if(ans > MOD) ans %= MOD;
    }
    cout << ans;
}
 
int main(){
    ios::sync_with_stdio(false);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    int q = 1;
    //cin >> q;
    for(int i = 1; i <= q; i++){
        solve();
    }
}