#include <bits/stdc++.h>

using namespace std;
vector<long> isprime(1002003);
vector<long> primes(100100);

vector<long> GetPrimeFactor(long n)
{
    int i;
    vector<long> answer;
    if (n<2) {
        return answer;      //return empty vector
    }
    while (n%2 == 0)
    {
        answer.push_back(2);
        n = n/2;
    }
    // n must be odd at this point.  So we can skip 
    // one element (Note i = i +2)
    for (int i = 1; primes[i] <= sqrt(n); i++) //watch for , n < 78498 
    {
        //cout << i << ":" << primes[i] << endl;
        // While i divides n, print i and divide n
        while ((n%primes[i] == 0) and (primes[i]!=0))
        {
            answer.push_back(primes[i]);
            n = n/primes[i];
        }
    }
    if (n>2) {answer.push_back(n);} //for case where n is prime to start with

    return answer;
}

void sieve()
{
    long k=0;
	primes[k++]=2;

	for(long i=3;i*i<=1002002;i+=2)
	{
		if(!isprime[i])
		{
			primes[k++]=i;
			for(long j=i*i;j<=1002002;j+=i+i)
				isprime[j] = 1;
		}
	}

	for(long i=1001;i<1002002;i+=2)
		if(!isprime[i])
			primes[k++]=i;
}

long getBreaks(long n)
{   
    if (n==1) return 1; if (n==2) return 3; if (n==3) return 4;
    long nBreaks=0;
    //nBreaks+=1; 
    nBreaks+=n; //add both 1 and n to the total
    //primes should be ordered
    vector<long> p = GetPrimeFactor(n);
    long runningProduct = 1;
    for (int i=0;i<p.size();i++) {
        runningProduct = runningProduct * p[i];
        nBreaks+=n/runningProduct;
    }
    //cout << "Number:" << n << " returns:" << nBreaks << endl;
    return nBreaks;
}

long longestSequence(vector <long> a) {
    long nSum=0;
    for (long i=0;i<a.size();i++) {
        long nBreaks = getBreaks(a[i]);
        nSum+=nBreaks;
    }
    return nSum;
}



int main() {
    sieve();
    int n;
    cin >> n;
    vector<long> a(n);
    for(long a_i = 0; a_i < n; a_i++){
       cin >> a[a_i];
    }
    long result = longestSequence(a);
    cout << result << endl;
    return 0;
}