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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <deque>
#include <iostream>
#include <iterator>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

#define FOR(i,k,n) for(int(i)=(k);(i)<(n);++(i))
#define rep(i,n) FOR(i,0,n)
#define all(v) begin(v),end(v)
#define debug(x) std::cerr<<#x<<": "<<x<<"\n"
#define debug2(x,y) std::cerr<<#x<<": "<<x<<", "<<#y<<": "<<y<<"\n"
#define debug3(x,y,z) std::cerr<<#x<<": "<<x<<", "<<#y<<": "<<y<<", "<<#z<<": "<<z<<"\n"

using ll=long long;
using vi=std::vector<int>;
using vvi=std::vector<vi>;
using vll=std::vector<ll>;
using vvll=std::vector<vll>;
template<typename T> std::vector<T> vec(int l,T e){return std::vector<T>(l,e);}
template<typename T> using MaxHeap=std::priority_queue<T>;
template<typename T> using MinHeap=std::priority_queue<T,std::vector<T>,std::greater<T> >;
template<typename T> inline void chmax(T&a,T b){a=std::max(a,b);}

template<typename T> std::ostream& operator<<(std::ostream&s,const std::vector<T>&v){
  int sz=v.size();s<<"\n";rep(i,sz){s<<v[i];if(i<sz-1){s<<"\t";}}s<<"\n";return s;}
template<typename T> std::ostream& operator<<(std::ostream&s,const std::vector<std::vector<T>>&v){
  for(auto&& w:v){s<<w;}return s;}
template<typename T> std::ostream& operator<<(std::ostream&s,const std::deque<T>&v){
  int sz=v.size();s<<"\n";rep(i,sz){s<<v[i];if(i<sz-1){s<<"\t";}}s<<"\n";return s;}
template<typename T> std::ostream& operator<<(std::ostream&s,const std::deque<std::deque<T>>&v){
  for(auto&& w:v){s<<w;}return s;}
template<typename T> std::ostream& operator<<(std::ostream&s, const std::set<T>&v){
  s<<"\n";for(auto&& elm:v){s<<elm<<"\t";}s<<"\n";return s;}

inline void scan(int&a){scanf("%d",&a);}
inline void scan(ll&a){scanf("%lld",&a);}
inline void scan(char&a){scanf(" %c",&a);}
inline void scan(double&a){scanf("%lf",&a);}
template<typename T>
inline void scan(std::vector<T>&v){for(auto&& sv:v){scan(sv);}}
template<typename First,typename...Args>
inline void scan(First&f,Args&...args){scan(f);scan(args...);}
inline void scan(std::string&s){char BUF[3000000];scanf(" %s",BUF);s=std::string(BUF);}
inline void print(int a){printf("%d\n",a);}
inline void print(ll a){printf("%lld\n",a);}
inline void print(double a){printf("%.12f\n",a);}
inline void print(std::string s){std::cout<<s<<"\n";}

using namespace std;

struct Prime {
  int max_check;
  int max_prime;
  int prime_number;
  vector<int> primes;
  map<int, int> prime_id;
  Prime() {};
  Prime(int m) {
    initialize(m);
  };
  void initialize(int m) {
    sieve(m);
    prime_number = (int)primes.size();
    max_check = m;
  }

  void sieve(int m) {
    max_prime = 2;
    deque<bool> is_prime(m+1, true);
    is_prime[0] = is_prime[1] = false;
    int i;
    for (i = 2; i <= m; ++i) {
      if (!is_prime[i]) {
        continue;
      }
      if (i * i > m) {
        break;
      }
      for (int j = i*2; j <= m; j += i) {
        is_prime[j] = false;
      }
      prime_id[i] = (int)primes.size();
      primes.push_back(i);
      max_prime = i;
    }
    for (int j = i; j <= m; ++j) {
      if (!is_prime[j]) {
        continue;
      }
      prime_id[j] = (int)primes.size();
      primes.push_back(j);
      max_prime = j;
    }
  }

  unordered_map<int, int> prime_factor(ll x) {
    unordered_map<int, int> ret;
    if ((ll)max_check * (ll)max_check < x) {
      return ret;
    }
    for (int i = 0; i < prime_number; ++i) {
      if (x == 1) {
        break;
      }
      while (x % primes[i] == 0) {
        x /= primes[i];
        ret[primes[i]] += 1;
      }
    }
    if (x != 1) {
      ret[x] += 1;
    }
    return ret;
  }
};

int mod = 1000000007;

ll po(ll k, ll x) {
  if (x == 0) {
    return 1;
  }
  if (x == 1) {
    return k % mod;
  }
  ll y = po(k, x/2);
  y = y * y % mod;
  if (x % 2 == 1) {
    y = y * k % mod;
  }
  return y;
}

int main() {
  int n;
  scan(n);
  vi a(n);
  scan(a);
  Prime pr(2000);
  unordered_map<int, int> lcm;
  rep (i, n) {
    auto pf = pr.prime_factor(a[i]);
    for (auto p : pf) {
      chmax(lcm[p.first], p.second);
    }
  }
  ll l = 1;
  for (auto p : lcm) {
    l = l * po(p.first, p.second) % mod;
  }

  ll ans = 0;
  rep (i, n) {
    ans = ans + po(a[i], mod-2);
    if (ans >= mod) {
      ans -= mod;
    }
  }
  ans = ans * l % mod;
  print(ans);

  return 0;
}