#include <iostream>
using namespace std;

#define MAX 100000010
int t[100000010];

int main(void){
	int n;

	while(cin>>n, n){
		int sum = 0;

		for(int i=1;i<n && i*i<=n;i++){
			if(n % i == 0){
				sum += i;
				if(i != n / i && n / i != n) sum += n / i;
			}
		}
		//		cout<<sum<<endl;

		cout<<(sum==n ? "perfect" : sum>n ? "abundant" : "deficient")<<" number\n";
	}

	return 0;
}
### 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 <iostream>
using namespace std;

#define MAX 100000010
int t[100000010];

int main(void){
	int n;

	while(cin>>n, n){
		int sum = 0;

		for(int i=1;i<n && i*i<=n;i++){
			if(n % i == 0){
				sum += i;
				if(i != n / i && n / i != n) sum += n / i;
			}
		}
		//		cout<<sum<<endl;

		cout<<(sum==n ? "perfect" : sum>n ? "abundant" : "deficient")<<" number\n";
	}

	return 0;
}
```