#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

int main(){
	int N;
	while (cin >> N, N) {
		vector<int> s;
		for (int i = 1; i < sqrt(N); i++) {
			if (N % i == 0) {s.push_back(i); s.push_back(N / i);}
		}
		int S = 0;
		for (int i = 0; i < s.size(); i++) {S += s[i];}
		S -= N;
		if (S - N) {
			if (S > N) cout << "abundant number" << endl;	
			else cout << "deficient number" << endl;
		} else {
			cout << "perfect 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 <iostream>
#include <vector>
#include <cmath>

using namespace std;

int main(){
	int N;
	while (cin >> N, N) {
		vector<int> s;
		for (int i = 1; i < sqrt(N); i++) {
			if (N % i == 0) {s.push_back(i); s.push_back(N / i);}
		}
		int S = 0;
		for (int i = 0; i < s.size(); i++) {S += s[i];}
		S -= N;
		if (S - N) {
			if (S > N) cout << "abundant number" << endl;	
			else cout << "deficient number" << endl;
		} else {
			cout << "perfect number" << endl;	
		}
	}
	return 0;
}
```