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
//------------------------------------------
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <bits/stdc++.h>
#include<queue>
#include <numeric>
using namespace std;
 
//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
 
//math
//-------------------------------------------
template<class T> inline T sqr(T x) {return x*x;}
 template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;
 
//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
 
//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define fore(i,a) for(auto &i:a)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);
 
//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))
 
//debug
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
 

LL gcd(LL a, LL b){
    if(b == 0) return a;
    return gcd(b,a%b);
}

LL lcm(LL a, LL b){
    LL g = gcd(a,b);
    return a/g *b;
}

int main(){
    long long N;
    cin >> N;
    long long A[N];
    REP(i,N) cin >> A[i];
    sort(A,A+N);
    long long limit=(int) pow(10,9)+7;
    long long sum=0;
    long long res=A[0];
    long long tmp=A[N-1];
    for(int i=0;i<N;++i){
        A[i]=A[i]%limit;
    }
    for(int i=0;i<N;++i){
        res=lcm(res,A[i]);
    }
    for(int i=0;i<N;++i){
        sum=(sum+res/A[i])%limit;
    }
    cout << sum;
}