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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define sz(x) (int)x.size()
#define int long long
typedef pair<int,int>pii;
const int mxN=1e4+10,mxA=1e6+10,M=1e9+7;
int mod(int a,int M){int ret=a%M;if(ret<0)ret+=M;return ret;}
int exp(int a,int b,int M){
    if(b==0LL)return 1LL;
    if(b==1LL)return mod(a,M);

    int t=exp(a,b/2,M);
    if(b&1)return mod(a*mod(t*t,M),M);
    return mod(t*t,M);
}

int n,x;
int v[mxN],p[mxA];
int32_t main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    cin>>n;
    for(int i=0;i<n;i++){
        cin>>v[i];
        x=v[i];
        for(int i=2;i*i<=x;i++){
            int cnt=0;
            if(x%i==0){
                while(x%i==0){
                    cnt++;
                    x/=i;
                }
                p[i]=max(p[i],cnt);
            }
        }
        if(x>1)p[x]=max(p[x],1LL);
    }
    int k=1LL;
    for(int i=2;i<=mxA;i++){
        while(p[i]--)k=mod(k*i,M);
    }
    int ans=0LL;
    for(int i=0;i<n;i++){
        ans=mod(ans+k*exp(v[i],M-2,M),M);
    }
    cout<<ans;

}