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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ll long long
#define pii pair<int,int>
#define pll pair<ll int,ll int>
#define pb push_back
#define sbit __builtin_popcount
#define vec vector <int>
#define ub upper_bound
#define lb lower_bound
#define bs binary_search
#define bg begin()
#define en end()
#define endl "\n"
#define ff first
#define ss second
#define all(x) x.bg,x.en
#define sz(a) (int)a.size()
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define sp(x) setprecision(x)
#define fill(a,b) memset(a,b,sizeof(a))
#define fr(i,n) for(int i=0;i<n;i++)
#define rp(i,a,b) for(int i=a;i<b;i++)
#define rr(i,a,b) for(int i=a;i>=b;i--)
#define ft(it,x) for(auto it : x)
#define dbg(x) cerr<<#x<<" "<<x<<endl;
#define dbga(arr,n) cerr<<#arr<<": "; rep(tt,0,n)cerr<<arr[tt]<<" "; cerr<<endl;
#define mapi map<int,int>
#define mapl map<ll int,ll int>
#define val_at(x) find_by_order(x)
#define idx_of(x) order_of_key(x)
#define scni(x) scanf("%d",&x)
#define prt(x) printf("%d\n",x)
#define test(t) int t;cin>>t;while(t--)
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define val_at(x) find_by_order(x)
#define idx_of(x) order_of_key(x)
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> myset;
const int mod=1e9+7;
const int MAX=1e6+5;
inline ll power(ll a,ll b)
{
    ll ans=1,temp=a;
    while(b!=0)
    {
        if(b&1)
        ans=ans*temp;
        ans=ans%mod;
        temp=temp*temp;
        temp=temp%mod;
        b/=2;
    }
    return ans;
}
ll int xx,xy,d1;
void exeu(ll int a1,ll int b1)
{
    if(b1==0)
    {
        xx=1;
        xy=0;
        d1=a1;
    }
    else
    {
        exeu(b1,a1%b1);
        ll int temp=xx;
        xx=xy;
        xy=temp-(a1/b1)*xy;
    }
}
ll int modinv(ll int A,ll int M)
{
    exeu(A,M);
    return (xx%M+M)%M;
}
ll int prime[MAX]; 
unordered_map<ll int,ll int> max_map; 
void sieve() 
{ 
    prime[0] = prime[1] = 1; 
    for (int i = 2; i < MAX; i++) { 
        if (prime[i] == 0) { 
            for (int j = i * 2; j < MAX; j += i) { 
                if (prime[j] == 0) { 
                    prime[j] = i; 
                } 
            } 
            prime[i] = i; 
        } 
    } 
} 
ll lcm(ll int ar[], int n) 
{ 
  
        for (int i = 0; i < n; i++) { 
            ll int num = ar[i]; 
            unordered_map<ll int,ll int> temp;
            while (num > 1) {
                ll int factor = prime[num];
                temp[factor]++;
                num /= factor; 
        } 
        for (auto it : temp) {
            max_map[it.first] = max(max_map[it.first], it.second); 
        } 
    } 
  
    ll ans = 1; 
    for (auto it : max_map) {
        ans = (ans * power(it.ff, it.ss)) % mod; 
    } 
  
    return ans; 
} 
int main()
{
    fast
    int n;
    cin>>n;
    ll int a[n];
    fr(i,n) cin>>a[i];
    sieve();
    ll int lm=lcm(a,n)%mod;
    ll ans=0;
    fr(i,n) {
        ll int ans1=(lm*modinv(a[i],mod))%mod;
        ans=(ans+ans1)%mod;
    }
    cout<<ans;
    return 0;
}