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.

Constraints

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

Examples

Input:
n = 18

Expected output:
expected = "abundant"

Input:
long long n = 18;

Expected output:
std::string expected = "abundant";

Input:
long n = 18;

Expected output:
string expected = "abundant";

Input:
n = 28

Expected output:
expected = "perfect"

Input:
long long n = 28;

Expected output:
std::string expected = "perfect";

Input:
long n = 28;

Expected output:
string expected = "perfect";

References

This challenge is adapted for CodeUtility. Follow the statement and constraints above.

• Perfect Numbers