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>//Carefully Crafted by hetp111
using namespace std;
#define int long long
#define double long double
#define all(v) (v).begin(),(v).end()
#define vi vector<int>
#define vvi vector<vi>
#define pii pair<int,int>
#define vii vector<pii>
#define MOD (1000000007)
#define PI acos(-1)
#define eps (1e-8)
#define INF (1e18)
#define FASTER ios_base::sync_with_stdio(0);cin.tie(0)
template<class A,class B>ostream&operator<<(ostream&out,const pair<A,B>&a){return out<<"("<<a.first<<","<<a.second<<")";}
template<class A>ostream&operator<<(ostream&out,const vector<A>&a){for(const A &it:a)out<<it<<" ";return out;}
template<class A,class B>istream&operator>>(istream&in,pair<A,B>&a){return in>>a.first>>a.second;}
template<class A>istream&operator>>(istream&in,vector<A>&a){for(A &i:a)in>>i;return in;}
//ifstream cinn("input.txt");ofstream coutt("output.txt");

int n;
int v[10001];

int lcm(int a,int b){
    return a*b/__gcd(a,b);
}

signed main(){
    FASTER;
    cin>>n;
    for(int i=0;i<n;i++)cin>>v[i];
    int lc=v[0];
    for(int i=1;i<n;i++){
        lc=lcm(lc,v[i]);
    }
    int ans=0;
    for(int i=0;i<n;i++){
        ans+=(lc/v[i]);
        if(ans>=MOD)ans-=MOD;
    }
    cout<<ans%MOD;
}