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

int main(){
     int N;
     while(cin>>N,N){
	  int sum=0;
	  int i;
	  for(i=(int)sqrt(N);i>0;i--){
	       if(N%i==0){
		    //cout<<i<<" ";
		    sum+=i;
		    if(N!=i*i)sum+=N/i;
	       }
	       //if(sum>N)break;
	  }
	  sum-=N;
	  if(sum>N){
	       cout<<"abundant number"<<endl;
	  }else if(sum==N){
	       cout<<"perfect number"<<endl;
	  }else{
	       cout<<"deficient number"<<endl;
	  }
     }
     return 0;
}
### Prompt

Create a solution in CPP for 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;
     while(cin>>N,N){
	  int sum=0;
	  int i;
	  for(i=(int)sqrt(N);i>0;i--){
	       if(N%i==0){
		    //cout<<i<<" ";
		    sum+=i;
		    if(N!=i*i)sum+=N/i;
	       }
	       //if(sum>N)break;
	  }
	  sum-=N;
	  if(sum>N){
	       cout<<"abundant number"<<endl;
	  }else if(sum==N){
	       cout<<"perfect number"<<endl;
	  }else{
	       cout<<"deficient number"<<endl;
	  }
     }
     return 0;
}
```