#include <iostream>
#include <cmath>

void check(int number) {
    int root_num = sqrt(number);
    int total = 0;

    for (int i = 1; i <= root_num; i++) {
        if (number % i == 0) {
            total += i;
            if (i != number / i) {
                total += (number / i);
            }
        }
    }
    total -= number;
    if (total == number) {
        std::cout << "perfect number" << std::endl;
    } else if (total < number) {
        std::cout << "deficient number" << std::endl;
    } else {
        std::cout << "abundant number" << std::endl;
    }

    return;
}

int main(void) {
    int number;
    while (1) {
        std::cin >> number;
        if (number == 0) {
            break;
        }
        check(number);
    }
    return 0;
}
### Prompt

Your challenge is to write a Cpp solution to the following problem:

Let S be the sum of divisors of an integer N excluding the number itself. When N = S, N is called a perfect number, when N> S, N is called a defendant number, and when N <S, N is called an abundant number. Create a program that determines whether a given integer is a perfect number, a missing number, or an abundant number.

Be careful not to exceed the program execution time.



Input

The input consists of a sequence of datasets. The number of datasets is 100 or less.

Each dataset consists of one row containing only the integer N (0 <N ≤ 100000000).

After the last dataset, there is a line marked 0 that marks the end of the input.

Output

For each dataset, print the string "` perfect number` "if the integer N is a perfect number," `deficient number`" if it is a missing number, or "` abundant number` "if it is an abundant number. ..

Example

Input

1
2
3
4
6
12
16
28
33550336
99999998
99999999
100000000
0


Output

deficient number
deficient number
deficient number
deficient number
perfect number
abundant number
deficient number
perfect number
perfect number
deficient number
deficient number
abundant number

### Response

```cpp
#include <iostream>
#include <cmath>

void check(int number) {
    int root_num = sqrt(number);
    int total = 0;

    for (int i = 1; i <= root_num; i++) {
        if (number % i == 0) {
            total += i;
            if (i != number / i) {
                total += (number / i);
            }
        }
    }
    total -= number;
    if (total == number) {
        std::cout << "perfect number" << std::endl;
    } else if (total < number) {
        std::cout << "deficient number" << std::endl;
    } else {
        std::cout << "abundant number" << std::endl;
    }

    return;
}

int main(void) {
    int number;
    while (1) {
        std::cin >> number;
        if (number == 0) {
            break;
        }
        check(number);
    }
    return 0;
}
```