Classify a Number by Its Divisors

Add the positive divisors of n that are strictly smaller than n. If their sum equals n, the number is perfect. If the sum is larger, it is abundant; if smaller, it is deficient. The number 1 has no proper positive divisors, so its divisor sum is 0.

Implement perfect_number(n).

Input

n: The positive integer to classify.

Output

Return perfect, abundant, or deficient as lowercase text.

Contraintes

1 ≤ n ≤ 1,000,000. Zero and negative values are excluded.

Exemples

Entrée :
n = 18

Sortie attendue :
expected = "abundant"

Entrée :
long long n = 18;

Sortie attendue :
std::string expected = "abundant";

Entrée :
long n = 18;

Sortie attendue :
string expected = "abundant";

Entrée :
n = 28

Sortie attendue :
expected = "perfect"

Entrée :
long long n = 28;

Sortie attendue :
std::string expected = "perfect";

Entrée :
long n = 28;

Sortie attendue :
string expected = "perfect";

Références

Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.

• Perfect Numbers