#include<iostream>

int main(){
	int n;
	while(std::cin>>n,n){
		int sum=0;
		for(int i=1;i*i<=n;i++){
			if(n%i==0){
				int j=n/i;
				if(i!=n)sum+=i;
				if(i!=j&&j!=n)sum+=j;
			}
		}
		if(sum==n)std::cout<<"perfect number"<<std::endl;
		else if(sum<n)std::cout<<"deficient number"<<std::endl;
		else if(sum>n)std::cout<<"abundant number"<<std::endl;
	}
	return 0;
}
### Prompt

Your challenge is to write 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>

int main(){
	int n;
	while(std::cin>>n,n){
		int sum=0;
		for(int i=1;i*i<=n;i++){
			if(n%i==0){
				int j=n/i;
				if(i!=n)sum+=i;
				if(i!=j&&j!=n)sum+=j;
			}
		}
		if(sum==n)std::cout<<"perfect number"<<std::endl;
		else if(sum<n)std::cout<<"deficient number"<<std::endl;
		else if(sum>n)std::cout<<"abundant number"<<std::endl;
	}
	return 0;
}
```