#include<iostream>
using namespace std;
int main()
{
  int n;
  int i;
  int sum;
  while(cin>>n){
    if(n==0) break;
    if(n==1){
      cout << "deficient number\n";
      continue;
    }
    sum=1;
    for(i=2;i*i<n;++i){
      if(n%i==0) sum+=(i+n/i);
    }
    if(sum==n) cout << "perfect number\n";
    else if(sum>n) cout <<"abundant number\n";
    else cout << "deficient number\n";
  }
  return 0;
}
### Prompt

Your challenge is to write 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>
using namespace std;
int main()
{
  int n;
  int i;
  int sum;
  while(cin>>n){
    if(n==0) break;
    if(n==1){
      cout << "deficient number\n";
      continue;
    }
    sum=1;
    for(i=2;i*i<n;++i){
      if(n%i==0) sum+=(i+n/i);
    }
    if(sum==n) cout << "perfect number\n";
    else if(sum>n) cout <<"abundant number\n";
    else cout << "deficient number\n";
  }
  return 0;
}
```