#include<iostream>
using namespace std;

int main(){

  int n;

  while(cin >> n, n){
    int k = -n+1;

    for(int i=2;i*i<n;i++){
      if( !(n%i) ) k += n/i + i;
    }

    if(!k && n != 1) cout << "perfect number";
    else if(k < 0 || n == 1) cout << "deficient number";
    else cout << "abundant number";
    cout << endl;
  }

}
### Prompt

Please provide a Cpp coded solution 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 k = -n+1;

    for(int i=2;i*i<n;i++){
      if( !(n%i) ) k += n/i + i;
    }

    if(!k && n != 1) cout << "perfect number";
    else if(k < 0 || n == 1) cout << "deficient number";
    else cout << "abundant number";
    cout << endl;
  }

}
```