#include <iostream>
#include <algorithm>
using namespace std;

int getS(int x) {
  if(x == 1) return 0;
  int sum = 1;
  int i;
  for(i = 2; i*i < x; ++i) {
    if(x % i == 0) sum += i + x/i;
  }
  if(i*i == x) sum += i;
  return sum;
}

int main() {
  int N;
  while(cin >> N && N) {
    int S = getS(N);
    if(N == S) {
      cout << "perfect number" << endl;
    } else if(N > S) {
      cout << "deficient number" << endl;
    } else if(N < S) {
      cout << "abundant number" << endl;
    }
  }
  return 0;
}
### Prompt

Generate 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 <algorithm>
using namespace std;

int getS(int x) {
  if(x == 1) return 0;
  int sum = 1;
  int i;
  for(i = 2; i*i < x; ++i) {
    if(x % i == 0) sum += i + x/i;
  }
  if(i*i == x) sum += i;
  return sum;
}

int main() {
  int N;
  while(cin >> N && N) {
    int S = getS(N);
    if(N == S) {
      cout << "perfect number" << endl;
    } else if(N > S) {
      cout << "deficient number" << endl;
    } else if(N < S) {
      cout << "abundant number" << endl;
    }
  }
  return 0;
}
```