#include<iostream>

using namespace std;

int main(void){
  int n,i;
  long long sum;


  while(1){
    cin >> n;
    if(n==0)break;

    sum=0;
    for(int i=1;i*i<n;i++){
      if(n%i==0){
	sum+=i;   
	if(i!=1)sum+=n/i;
      }
    }
    if(n==sum)cout << "perfect number" << endl;
    else if(n<sum)cout << "abundant number" << endl;
    else cout << "deficient number" << endl;
  }

  return 0;
}
### Prompt

Please create a solution in Cpp 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>

using namespace std;

int main(void){
  int n,i;
  long long sum;


  while(1){
    cin >> n;
    if(n==0)break;

    sum=0;
    for(int i=1;i*i<n;i++){
      if(n%i==0){
	sum+=i;   
	if(i!=1)sum+=n/i;
      }
    }
    if(n==sum)cout << "perfect number" << endl;
    else if(n<sum)cout << "abundant number" << endl;
    else cout << "deficient number" << endl;
  }

  return 0;
}
```