#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<math.h>
using namespace std;
  
int main(){
	long int n,m,i,j,k;
	while(1){
		m=0;j=0;k=0;
		cin>>n;
		if(n==0){break;}
		
		for(i=1;i<=sqrt((double)n);i++){
			if(n%i==0){
				if(i==1){
					m++;
				}else if(i==sqrt((double)n)){
					m+=i;
				}else{
					m+=i+(n/i);
				}
			}
		}
		if(n==1){
			cout<<"deficient number"<<endl;
		}else if(n==m){
			cout<<"perfect number"<<endl;
		}else if(n<m){
			cout<<"abundant number"<<endl;
		}else{
			cout<<"deficient number"<<endl;
		}
	}
    return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<math.h>
using namespace std;
  
int main(){
	long int n,m,i,j,k;
	while(1){
		m=0;j=0;k=0;
		cin>>n;
		if(n==0){break;}
		
		for(i=1;i<=sqrt((double)n);i++){
			if(n%i==0){
				if(i==1){
					m++;
				}else if(i==sqrt((double)n)){
					m+=i;
				}else{
					m+=i+(n/i);
				}
			}
		}
		if(n==1){
			cout<<"deficient number"<<endl;
		}else if(n==m){
			cout<<"perfect number"<<endl;
		}else if(n<m){
			cout<<"abundant number"<<endl;
		}else{
			cout<<"deficient number"<<endl;
		}
	}
    return 0;
}
```