#include <iostream>

using namespace std;

int main(void){

  long long n;

  while( cin >> n , n ){

    if( n == 1 ){
      cout << "deficient number" << endl;
      continue;
    }

    long long total = 1;
    long long a = 0;

    for( int i = 2 ; a != i && i < n/2 ; i++ ){
      if( n % i == 0 ){
	total += i;
	total += ( n / i );
	a = n / i;
	if( a == i ){
	  total -= i;
	  break;
	}
      }
      if( total > n ) break;
    }
    //cout << "a = " << a << endl;
    //cout << total << "    ";
    if( n == total ) cout << "perfect number" << endl;
    if( n < total ) cout << "abundant number" << endl;
    if( n > total ) cout << "deficient 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>

using namespace std;

int main(void){

  long long n;

  while( cin >> n , n ){

    if( n == 1 ){
      cout << "deficient number" << endl;
      continue;
    }

    long long total = 1;
    long long a = 0;

    for( int i = 2 ; a != i && i < n/2 ; i++ ){
      if( n % i == 0 ){
	total += i;
	total += ( n / i );
	a = n / i;
	if( a == i ){
	  total -= i;
	  break;
	}
      }
      if( total > n ) break;
    }
    //cout << "a = " << a << endl;
    //cout << total << "    ";
    if( n == total ) cout << "perfect number" << endl;
    if( n < total ) cout << "abundant number" << endl;
    if( n > total ) cout << "deficient number" << endl;

  }
  return 0;
}
```