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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pii;
typedef vector<pii> vpii;
typedef vector<vpii> vvpii;

#define mt make_tuple
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sqr(x) ((ll)(x) * (x))

const ll MOD = 1e9+7LL;
const int MAXA = 1e6+1;

ll cinioci[1001];

ll modPow(ll a, ll n, ll m)
{
    if(n == 0)
        return 1LL;
    ll pola = modPow(a, n/2, m)%m;
    ll celo = (pola*pola)%m;

    if(n % 2 == 0)
    {
        return celo;
    }
    else
    {
        return (celo*a)%m;
    }
}

void fact(int n)
{
    for(int i = 2; i * i <= n; ++i)
    {
        int brojac = 0;

        while(n % i == 0)
        {
            n /= i;
            ++brojac;
        }

        cinioci[i] = max(cinioci[i], (ll)brojac);
    }

    if(n > 1)
    {
        cinioci[n] = max(cinioci[n], 1LL);
    }
}

int main()
{
    int n;
    cin >> n;
    vector<ll> v(n);

    memset(cinioci, 0, sizeof(cinioci));

    for(auto &z : v)
    {
        cin >> z;
        fact(z);
    }

    ll nzs = 1LL;

    for(int i = 2; i <= 1000; ++i)
    {
        nzs = (nzs*modPow(i, cinioci[i], MOD))%MOD;
    }

    //cout << nzs << '\n';

    ll resenje = 0LL;

    for(int i = 0; i < n; ++i)
    {
        resenje = (resenje + (nzs*modPow(v[i], MOD-2, MOD))%MOD)%MOD;
    }

    cout << resenje%MOD << '\n';

    return 0;
}