#include<iostream>
#include<cmath>
using namespace std;
int main (){

  int n,a,b;

  while(1){

    a=0;
    
    cin >>n;
    
    if( n == 0 )break;

    for(int i=1; i <=  sqrt(n) ; i++){
      if( n%i == 0){
	if( i == n/i){
	  a+=i;
	}else{
	  a+=i+n/i;
	}      
      }
    }
    
    
    b=a-n;

    if( n == 1 ){
      b=0;
    }

    if( n == b ){
      cout <<"perfect number"<<endl;
    }else if( n > b ){
      cout <<"deficient number"<<endl;
    }else if( n < b ){
      cout <<"abundant 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>
#include<cmath>
using namespace std;
int main (){

  int n,a,b;

  while(1){

    a=0;
    
    cin >>n;
    
    if( n == 0 )break;

    for(int i=1; i <=  sqrt(n) ; i++){
      if( n%i == 0){
	if( i == n/i){
	  a+=i;
	}else{
	  a+=i+n/i;
	}      
      }
    }
    
    
    b=a-n;

    if( n == 1 ){
      b=0;
    }

    if( n == b ){
      cout <<"perfect number"<<endl;
    }else if( n > b ){
      cout <<"deficient number"<<endl;
    }else if( n < b ){
      cout <<"abundant number"<<endl;
    }

  }
    


  
  return 0;
}
```