#include <iostream>
using namespace std;

int main(void) {
  long long N, sum;

  while (1) {
    cin >> N;
    if (N == 0) break;
    sum = 1;

    for (long long i = 2; i*i <= N; i++) {
      if (N % i == 0) {
        long long tmp = N / i;
        sum += i;
        if (i != tmp) {
          sum += tmp;
        }
      }
      if (sum > N) {
        cout << "abundant number" << endl;
        break;
      }
    }

    if (N == 1 || sum < N) {
      cout << "deficient number" << endl;
    } else if (sum == N) {
      cout << "perfect number" << endl;
    }
  }

  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(void) {
  long long N, sum;

  while (1) {
    cin >> N;
    if (N == 0) break;
    sum = 1;

    for (long long i = 2; i*i <= N; i++) {
      if (N % i == 0) {
        long long tmp = N / i;
        sum += i;
        if (i != tmp) {
          sum += tmp;
        }
      }
      if (sum > N) {
        cout << "abundant number" << endl;
        break;
      }
    }

    if (N == 1 || sum < N) {
      cout << "deficient number" << endl;
    } else if (sum == N) {
      cout << "perfect number" << endl;
    }
  }

  return 0;
}
```