#include<iostream>
#include<map>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
int main(){
	int n;
	while(1){
		cin >> n;
		if(n==0) break;
		int sum,hoge=n;
		n<2?sum=0:sum=1;
		for(int i=2;i< hoge && sum < n;i++){
			if(n%i == 0){      //約数
				hoge= n/i;     //n = i * hoge
				if(i!=hoge){   
					sum += i+hoge;
				}else{
					sum += i;
				}
			}
		}
	if(sum==n)
		cout << "perfect number" << endl;
	else if(sum < n)
		cout << "deficient number" << endl;
	else
		cout << "abundant number" << endl;
	}
}
### Prompt

In Cpp, your task is to solve 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<map>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
int main(){
	int n;
	while(1){
		cin >> n;
		if(n==0) break;
		int sum,hoge=n;
		n<2?sum=0:sum=1;
		for(int i=2;i< hoge && sum < n;i++){
			if(n%i == 0){      //約数
				hoge= n/i;     //n = i * hoge
				if(i!=hoge){   
					sum += i+hoge;
				}else{
					sum += i;
				}
			}
		}
	if(sum==n)
		cout << "perfect number" << endl;
	else if(sum < n)
		cout << "deficient number" << endl;
	else
		cout << "abundant number" << endl;
	}
}
```