#include<stdio.h>
#include<iostream>
#include<fstream>
#include<stack>
#include<math.h>

using namespace std;



int main(){

	int n;
	while (cin >> n, n){
		if (n < 2){
			cout << "deficient number" << endl;
			continue;
		}

		int s = 1;
		for (int i = 2; i <= sqrt(n); i++){
			if (n%i == 0 && i == sqrt(n) && n / sqrt(n) == sqrt(n)){
				s += i;
			}
			else if (n%i == 0){
				s += i;
				s += n / i;
			}
		}

		if (n == s)cout << "perfect number" << endl;
		if (n > s)cout << "deficient number" << endl;
		if (n < s)cout << "abundant number" << 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<stdio.h>
#include<iostream>
#include<fstream>
#include<stack>
#include<math.h>

using namespace std;



int main(){

	int n;
	while (cin >> n, n){
		if (n < 2){
			cout << "deficient number" << endl;
			continue;
		}

		int s = 1;
		for (int i = 2; i <= sqrt(n); i++){
			if (n%i == 0 && i == sqrt(n) && n / sqrt(n) == sqrt(n)){
				s += i;
			}
			else if (n%i == 0){
				s += i;
				s += n / i;
			}
		}

		if (n == s)cout << "perfect number" << endl;
		if (n > s)cout << "deficient number" << endl;
		if (n < s)cout << "abundant number" << endl;

	}

	return 0;

}
```