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;
using ll=long long;
using ull=unsigned long long;
using pii=pair<int,int>;

#define INF LONG_MAX
#define MOD 1000000007
#define rng(a) a.begin(),a.end()
#define rrng(a) a.end(),a.begin()
#define rep(i,N) for(int i=0;i<N;i++)

template<ll mod=MOD>
ll modpow(ll x,ll n){
  if(n==0)return 1;
  if(n%2)return ((x%mod)*(modpow(x,n-1)%mod))%mod;
  else {
    ll half=modpow(x,n/2)%mod;
    return (half*half)%mod;
  }
}

#define int ll

signed main(){
  ios::sync_with_stdio(false);
  cin.tie(0);

  int N;
  cin>>N;
  vector<ll>A(N);
  for(int i=0;i<N;i++)cin>>A[i];

  vector<ll>B=A;
  vector<map<int,ll>>vec(N);
  for(int i=0;i<N;i++){
    for(int j=2;j*j<=A[i];j++){
      while(A[i]%j==0)A[i]/=j,vec[i][j]=(vec[i][j]+1)%MOD;
    }
    if(N!=1)vec[i][A[i]]=1;
  }

  map<int,ll>GCD;
  for(int i=0;i<N;i++){
    for(auto it:vec[i]){
      GCD[it.first]=max(GCD[it.first],it.second);
    }
  }

  ll gcd=1;
  for(auto it:GCD)gcd=((gcd%MOD)*modpow(it.first,it.second))%MOD;
  gcd%=MOD;

  for(int i=0;i<N;i++)B[i]%=MOD;
  ll ans=0;
  gcd%=MOD;
  for(int i=0;i<N;i++){
    ans=(((modpow(B[i],MOD-2)*gcd)%MOD)+ans%MOD)%MOD;
  }
  cout<<ans%MOD<<endl;

  return 0;
}