//Copyright © 2017 Snehil Santhalia
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef pair<ll,ll> ii;
typedef vector<ii> vii;
typedef vector<vii> vvii;

#define INF 1e18
#define loop(i,a,b) for(ll i=(ll)a;i<=(ll)b;i++)
#define forit(i, a) for ( __typeof( (a).begin() ) i = (a).begin(); i != (a).end(); i++ )
#define mem(a, v) memset(a, v, sizeof a)
#define pb push_back
#define mp make_pair
#define MOD 1000000007
#define MAX(a,b) (((a)>(b))?(a):(b))
#define MIN(a,b) (((a)<(b))?(a):(b))
#define left(x) x<<1
#define right(x) (x<<1)|1
#define PI acos(-1.0)
#define EPS 1e-9

int T = 1, N;
vi factors[1000505];
ll ans[1000505], no;
bitset<1000505> bs;
vi primes;

void sieve(){
    ll n=1000005;
    bs.set();
    bs[0]=bs[1]=0;
    loop(i,2,n)
        if(bs[i]){
            primes.pb(i);
            for(ll j=i;j<=n;j+=i) {
                bs[j]=0;
                factors[j].pb(i);
            }
        }
}

bool isPrime(ll N){
    if(N<=1000005) return bs[N];
    else {
        for(int i=0;i<(int)primes.size();i++)
            if(!(N%primes[i]))
                return false;
        return true;
    }
}

void solve(){
	cin >> N;
    ll tot = 0, j;

    loop(i,1,N) {
        cin >> no;
        while(no > 1) {
            if(isPrime(no)) {tot += no+1; break;}
            for(j = 0; j<(int)primes.size(); j++)
                if(no % primes[j] == 0) {
                    tot += no;
                    no /= primes[j];
                    break;
                }
        }
        if(no == 1) tot++;
    }
    cout << tot << endl;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    
    //cin>>T;
    sieve();
    while(T--){
    	solve();
    }
    return 0;
}