#include <iostream>
#include <algorithm>
#include <string>

#define rep(i, n) for(int i = 0; i < (n); i++)

using namespace std;

int main(){
	int n;
	while(cin >> n, n){
		int sum = n!=1;
		for(int i = 2; i*i <= n; i++){
			if(n%i==0){
				sum += i;
				if(i*i != n) sum += n/i;
			}
		}
		string s;
		if(sum == n){
			s = "perfect number";
		}else if(sum < n){
			s = "deficient number";
		}else{
			s = "abundant number";
		}
		cout << s << endl;
	}


	return 0;
}
### Prompt

Generate 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 <algorithm>
#include <string>

#define rep(i, n) for(int i = 0; i < (n); i++)

using namespace std;

int main(){
	int n;
	while(cin >> n, n){
		int sum = n!=1;
		for(int i = 2; i*i <= n; i++){
			if(n%i==0){
				sum += i;
				if(i*i != n) sum += n/i;
			}
		}
		string s;
		if(sum == n){
			s = "perfect number";
		}else if(sum < n){
			s = "deficient number";
		}else{
			s = "abundant number";
		}
		cout << s << endl;
	}


	return 0;
}
```