#include <iostream>
using namespace std;

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

int main() {
  int n;
  
  while( cin >> n, n ) {
    int perfectNumber = 0;
    
    for(int i=1; i*i<n; i++) {
      if( n % i == 0 ) {
	perfectNumber += n/i;
	perfectNumber += i;
      }
    }
    perfectNumber -= n;
    if( perfectNumber < n ) cout << "deficient number" << endl;
    else if( perfectNumber == n ) cout << "perfect number" << endl;
    else cout << "abundant number" << endl;
  }
  
  return 0;
}
```