#include<iostream>
#include<cmath>
#define loop(i,a,b) for(int i=a;i<b;i++)
#define rep(i,a) loop(i,0,a)
using namespace std;

int main(){
	long long s,n;
	while(cin>>n,n){
		s=1;
		loop(i,2,sqrt(n))if(n%i==0){s+=i;s+=(n/i);}
		if(n==1)cout<<"deficient number"<<endl;
		else if(n>s)cout<<"deficient number"<<endl;
		else if(n<s)cout<<"abundant number"<<endl;
		else cout<<"perfect number"<<endl;
	}
	return 0;
}
### Prompt

Your task is to create 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>
#define loop(i,a,b) for(int i=a;i<b;i++)
#define rep(i,a) loop(i,0,a)
using namespace std;

int main(){
	long long s,n;
	while(cin>>n,n){
		s=1;
		loop(i,2,sqrt(n))if(n%i==0){s+=i;s+=(n/i);}
		if(n==1)cout<<"deficient number"<<endl;
		else if(n>s)cout<<"deficient number"<<endl;
		else if(n<s)cout<<"abundant number"<<endl;
		else cout<<"perfect number"<<endl;
	}
	return 0;
}
```