#include<cstdio>
#include<iostream>
#include<cmath>

using namespace std;
 
int main(void){
 
    int N,S,f,cnt;
 
    while(1){
        S=0;
		cnt=0;
        cin>>N;
        if(N==0) break;
        if(N%2==0) f=1;
        else f=2;
        for(int i=1;i*i<N;i+=f){
            if(N%i==0){
				S+=i+(N/i);
			}
        }
		S-=N;
		if(sqrt((double)N)==(double)(int)sqrt((double)N)) S-=(int)sqrt((double)N);
        if(N==S) cout<<"perfect number"<<endl;
        if(N>S)  cout<<"deficient number"<<endl;
        if(N<S)  cout<<"abundant number"<<endl;
    }
     
    return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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<cstdio>
#include<iostream>
#include<cmath>

using namespace std;
 
int main(void){
 
    int N,S,f,cnt;
 
    while(1){
        S=0;
		cnt=0;
        cin>>N;
        if(N==0) break;
        if(N%2==0) f=1;
        else f=2;
        for(int i=1;i*i<N;i+=f){
            if(N%i==0){
				S+=i+(N/i);
			}
        }
		S-=N;
		if(sqrt((double)N)==(double)(int)sqrt((double)N)) S-=(int)sqrt((double)N);
        if(N==S) cout<<"perfect number"<<endl;
        if(N>S)  cout<<"deficient number"<<endl;
        if(N<S)  cout<<"abundant number"<<endl;
    }
     
    return 0;
}
```