#include <iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
int main(void){
    int N;
    while(1){
        cin>>N;
        if(N==0){
            break;
        }
        if(N==1){
            cout<<"deficient number"<<endl;
        }
        else{
        int S=0;
        for(int i=1;i<=sqrt(N);i++){
            if(N%i==0){
            S+=i;
            if(i!=(N/i))
            S+=(N/i);
            }
        }
        S=S-N;
        if(N==S){
            cout<<"perfect number"<<endl;
        }
        else if(N<S){
            cout<<"abundant number"<<endl;
        }
        else{
            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>
#include<stdio.h>
#include<math.h>
using namespace std;
int main(void){
    int N;
    while(1){
        cin>>N;
        if(N==0){
            break;
        }
        if(N==1){
            cout<<"deficient number"<<endl;
        }
        else{
        int S=0;
        for(int i=1;i<=sqrt(N);i++){
            if(N%i==0){
            S+=i;
            if(i!=(N/i))
            S+=(N/i);
            }
        }
        S=S-N;
        if(N==S){
            cout<<"perfect number"<<endl;
        }
        else if(N<S){
            cout<<"abundant number"<<endl;
        }
        else{
            cout<<"deficient number"<<endl;            
        }
        }
    }
    return 0;
    
}


```