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 <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
#include <complex>
#include <array>
#include <iomanip>
using namespace std;

#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<long double> VD;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef vector<VD> VVD;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

template<typename T> void chmin(T &a, T b) { if (a > b) a = b; }
template<typename T> void chmax(T &a, T b) { if (a < b) a = b; }

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }
#define INF 1LL<<60

ll mod = 1e9 + 7;

long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

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

template< typename T >
T mod_pow(T x, T n, const T &p) {
  T ret = 1;
  while(n > 0) {
    if(n & 1) (ret *= x) %= p;
    (x *= x) %= p;
    n >>= 1;
  }
  return ret;
}

map< ll, ll > prime_factor(ll n) {
  map< ll, ll > ret;
  for(ll i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}

int main() {
    int N;
    cin >> N;

    vector<ll> A(N);
    map<ll, ll> lcm;
    vector<map<ll, ll> > A_factors(N);
    ll ans = 0;
    REP(i, N) {
        cin >> A[i];
        map<ll, ll> A_factor = prime_factor(A[i]);
        A_factors[i] = A_factor;
        for(auto p = A_factor.begin(); p!=A_factor.end(); ++p)
            lcm[p->first] = max(lcm[p->first], p->second);
    }

    REP(i, N) {
        map<ll, ll> B_factor;
        ll b = 1;
        for(auto p = lcm.begin(); p != lcm.end(); ++p) {
            ll n = p->second - A_factors[i][p->first];
            b *= mod_pow(p->first, n, mod);
            b %= mod;
        }

        ans += b;
        ans %= mod;
    }

    cout << ans << endl;

    return 0;
}