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).

Examples

Input

3
2 3 4


Output

13


Input

5
12 12 12 12 12


Output

5


Input

3
1000000 999999 999998


Output

996989508
#include<bits/stdc++.h>
using namespace std;
static const int64_t mod=1000000007;
int64_t f(int64_t x,int64_t y){
if(y==0)
  return 1;
  else if(y%2==0)
    return (f(x,y/2)*f(x,y/2))%mod;
  else
    return (f(x,y/2)*((x*f(x,y/2))%mod))%mod;
}
int main(){
int N;
  cin>>N;
  vector<int64_t>A(N);
  for(int i=0;i<N;i++)
    cin>>A.at(i);
  map<int64_t,int>M;
  for(int i=0;i<N;i++){
  int64_t n=A.at(i);
    int64_t X=n;
    for(int64_t j=2;j*j<=n;j++){
    if(X%j==0){
    int a=0;
      while(X%j==0){X/=j; a++;}
    
    if(M.count(j)){
    if(M.at(j)<a) M.at(j)=a;
    }else
      M[j]=a;
    }
    
    }if(X!=1){ if(!M.count(X)){
    
      M[X]=1;
    }}
      
    }int64_t lcm=1;
  for(auto p:M){
  auto k=p.first;
    	auto v=p.second;
    lcm=(lcm*f(k,v))%mod;
  }int64_t ans=0;
  for(int i=0;i<N;i++){
  int64_t x=A.at(i);
    ans=(ans+lcm*f(x,mod-2))%mod;
  }cout<<ans<<endl;
   return 0;
}

