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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <numeric>
#include <map>
#include <iomanip>
#include <limits.h>
#include <queue>
#include <numeric>
using namespace std;    
typedef long long ll;
typedef pair<ll, ll> pll;
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
const bool DEBUG = false;

const ll MOD = 1000000007;

// mod. m での a の逆元 a^{-1} を計算する
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

ll gcd(ll a,ll b){
    if(a<b)swap(a,b);
    if(a%b==0)return b;
    return gcd(a%b,b);
}
ll lcm(ll a,ll b){
    return a/gcd(a,b)*b;
}


// a ÷ b mod. MOD を計算してみる
int main() {

    ll n;
    cin >> n;
 
    vector<ll> v(n);
    for(int i = 0; i < n; i++) cin >> v[i];
 
    ll LSM = 1;
    for(int i = 0; i < n; i++){
        LSM = lcm( LSM, v[i] );
        LSM %= MOD;
    }

    ll ans = 0;
    for(int i = 0; i < n; i++){
        ans += (LSM * modinv(v[i], MOD) );
        ans %= MOD;
    }
    cout << ans << endl;
}