#include <iostream>

using namespace std;

int main()
{
    int N;

    while (cin >> N, N) {
        int S = 0;

        for (int i = 1; i * i <= N; ++i) {
            if (N % i == 0 && i != N) {
                S += i;
                if (i != N / i && N / i != N)
                    S += N / i;
            }
        }

        if (N == S)
            cout << "perfect number\n";
        else if (N > S)
            cout << "deficient number\n";
        else
            cout << "abundant number\n";
    }

    return 0;
}
### Prompt

Please formulate 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>

using namespace std;

int main()
{
    int N;

    while (cin >> N, N) {
        int S = 0;

        for (int i = 1; i * i <= N; ++i) {
            if (N % i == 0 && i != N) {
                S += i;
                if (i != N / i && N / i != N)
                    S += N / i;
            }
        }

        if (N == S)
            cout << "perfect number\n";
        else if (N > S)
            cout << "deficient number\n";
        else
            cout << "abundant number\n";
    }

    return 0;
}
```