#include <iostream>
using namespace std;

int main(void){

  int N,S;
  int sum;

  while(1){

    cin >> N;

    if( N == 0 ) break;

      sum = 0;

      for( int j = 1 ; j*j <= N; j++ ){
	
	if( N % j == 0 ){ 
	  sum += j;
	  if( j*j != N ) sum += N/j;
           
	}
    }
      
      sum -= N;

    if( N < sum ) cout << "abundant number" << endl;
    else if( N > sum ) cout << "deficient number" << endl;
    else cout << "perfect number" << endl;

  }

  return 0;

}
### 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>
using namespace std;

int main(void){

  int N,S;
  int sum;

  while(1){

    cin >> N;

    if( N == 0 ) break;

      sum = 0;

      for( int j = 1 ; j*j <= N; j++ ){
	
	if( N % j == 0 ){ 
	  sum += j;
	  if( j*j != N ) sum += N/j;
           
	}
    }
      
      sum -= N;

    if( N < sum ) cout << "abundant number" << endl;
    else if( N > sum ) cout << "deficient number" << endl;
    else cout << "perfect number" << endl;

  }

  return 0;

}
```