When n is a prime number 2,3,5... 19, the result has to be copy paste one by one and result in n operations. For any prime divisor of n, this is also true. So this question actually becomes calculate the sum of its prime divisors. Attach my code your reference.

```
public int minSteps(int n) {
Set<Integer> set = new HashSet<>();
boolean[] notPrime = new boolean[n+1];
for (int i = 2; i * i <= n; i++) {
if (notPrime[i]) {
continue;
}
set.add(i);
for (int j = i + i; j <= n; j+= i) {
notPrime[j] = true;
}
}
return minSteps(n, set);
}
private int minSteps(int n, Set<Integer> set) {
if (n == 0 || n == 1) {
return 0;
}
Set<Integer> remove = new HashSet<>();
for (int v : set) {
if (n % v == 0) {
set.removeAll(remove); // for any non dividable prime, remove it
return v + minSteps(n/v, set);
} else {
remove.add(v);
}
}
return n;
}
```