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: Runtime Error)

# include <bits/stdc++.h>
# define ll int64_t
# define rep(i,n) for(ll i=0;i<n;i++)
# define rrep(i,n) for(ll i=1;i<=n;i++)
# define ALL(x) (x).begin(), (x).end()
# define SZ(x) ((int)(x).size())
# define pb push_back
# define mod 1000000007

using namespace std;

bool compare_by_b(pair<string,ll> a,pair<string,ll> b){
    if(a.second != b.second) return a.second<b.second;
    else return a.first<b.first;
}

bool my_compare(pair<string,ll> a,pair<string,ll> b){
    
    if(a.first != b.first) return a.first<b.first;
    if(a.second != b.second) return a.second>b.second;
    else return true;
}

ll factorial(ll n){
    ll x=1;
    rrep(i,n) x*=i;
    return x;
}

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

ll lcm(ll a,ll b)
{
   return a/gcd(a,b)*b;
}

void factoring(ll n, map<ll, ll>& mp) {
    for (ll i = 2; i * i <= n; i++) {
        while (n % i == 0) {
            mp[i]++;
            n /= i;
        }
    }
    if (n != 1) mp[n]++;
}


int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    ll n,sum=0;
    cin>>n;
    vector<ll> a(n);
    rep(i,n) cin>>a[i];
    
    ll first=lcm(a[0],a[1]);
    rrep(i,n-2){
        first=lcm(first,a[i+1]);
    }
    rep(i,n){
        sum+=first/a[i];
        sum%=mod;
    }
    cout<<sum<<endl;
    return 0;
}