#include <iostream>
#include <cstdlib>

using namespace std;

int main() {
  
  int n;
  while( 1 ) {
    cin >> n;
    if( !n ) { break; }

    if( n == 1 ) {
      cout << "deficient number" << endl;
      continue;
    }
    int sum = 1;
    for( int i = 2; i * i < n; i++ ) {
      if( n % i == 0 ) { 
        sum += i;
        if( i * i != n ) {
          sum += n / i;
        }

        if( sum > n ) { break; }
      }
    }

    cout << (
      ( sum  < n ) ? "deficient number" :
      ( sum == n ) ? "perfect number"   :
                     "abundant number"
    ) << endl;
  }

  return EXIT_SUCCESS;
}
### 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 <cstdlib>

using namespace std;

int main() {
  
  int n;
  while( 1 ) {
    cin >> n;
    if( !n ) { break; }

    if( n == 1 ) {
      cout << "deficient number" << endl;
      continue;
    }
    int sum = 1;
    for( int i = 2; i * i < n; i++ ) {
      if( n % i == 0 ) { 
        sum += i;
        if( i * i != n ) {
          sum += n / i;
        }

        if( sum > n ) { break; }
      }
    }

    cout << (
      ( sum  < n ) ? "deficient number" :
      ( sum == n ) ? "perfect number"   :
                     "abundant number"
    ) << endl;
  }

  return EXIT_SUCCESS;
}
```