#include<iostream>

using namespace std;

int main(void)
{

  int n;
  int sum;
  char str[][100] = {
    {"deficient number\0"},
    {"abundant number\0"},
    {"perfect number\0"}
  };

  while(cin >> n && n){
    sum = 1;
    for(int i = 2; i*i < n; i++){
      if(!(n%i)) sum += i + n / i;
    }
    if(n!=1){
      if(sum == n){ cout << str[2] << endl; }
      if(sum >  n){ cout << str[1] << endl; }
      if(sum <  n){ cout << str[0] << endl; }
    }
    else          { cout << str[0] << endl; }
  }

  return 0;
}
### Prompt

Create a solution in Cpp for 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;
  int sum;
  char str[][100] = {
    {"deficient number\0"},
    {"abundant number\0"},
    {"perfect number\0"}
  };

  while(cin >> n && n){
    sum = 1;
    for(int i = 2; i*i < n; i++){
      if(!(n%i)) sum += i + n / i;
    }
    if(n!=1){
      if(sum == n){ cout << str[2] << endl; }
      if(sum >  n){ cout << str[1] << endl; }
      if(sum <  n){ cout << str[0] << endl; }
    }
    else          { cout << str[0] << endl; }
  }

  return 0;
}
```