#include <bits/stdc++.h>
using namespace std;

int main() {
	int N;
	
	while (cin >> N, N) {	
		int sum = 1;
		
		for (int x = 2; x * x <= N; ++x) {
			if (N % x == 0) {
				sum += x;
				if (x != N / x) sum += N / x;
			}
		}
		
		if (N == 1 || sum < N) cout << "deficient number" << endl;
		else cout << (sum == N ? "perfect number" : "abundant number") << endl;
	}
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <bits/stdc++.h>
using namespace std;

int main() {
	int N;
	
	while (cin >> N, N) {	
		int sum = 1;
		
		for (int x = 2; x * x <= N; ++x) {
			if (N % x == 0) {
				sum += x;
				if (x != N / x) sum += N / x;
			}
		}
		
		if (N == 1 || sum < N) cout << "deficient number" << endl;
		else cout << (sum == N ? "perfect number" : "abundant number") << endl;
	}
}
```