#include <iostream>
using namespace std;

int main(){
	int n, s;

	while(true){
		cin >> n;
		if(n == 0){
			break;
		}
		s = 1;

		if(n == 1 || n == 2 || n == 3){
			cout << "deficient number" << endl;
			continue;
		}
		for(int i = 2; i * i <= n; i++){
			if(n % i == 0){
				s = s + i;
				if(i != n / i){
					s = s + n / i;
				}
			}
		}
		if(n == s){
			cout << "perfect number" << endl;
		}
		else if(n > s){
			cout << "deficient number" << endl;
		}
		else if(n < s){
			cout << "abundant number" << endl;
		}

	}
}
### Prompt

Please formulate 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>
using namespace std;

int main(){
	int n, s;

	while(true){
		cin >> n;
		if(n == 0){
			break;
		}
		s = 1;

		if(n == 1 || n == 2 || n == 3){
			cout << "deficient number" << endl;
			continue;
		}
		for(int i = 2; i * i <= n; i++){
			if(n % i == 0){
				s = s + i;
				if(i != n / i){
					s = s + n / i;
				}
			}
		}
		if(n == s){
			cout << "perfect number" << endl;
		}
		else if(n > s){
			cout << "deficient number" << endl;
		}
		else if(n < s){
			cout << "abundant number" << endl;
		}

	}
}
```