#include<iostream>
#include<math.h>
using namespace std;

long long int pre(long long int n)
{
  if(n<2)
    return 0;
  long long int c=1;
  for(int i=2;i<=sqrt(n);i++)
    if(n%i==0)
      {
	c+=i;
	if(i!=n/i)
	  c+=n/i;
      }
  return c;
}

int main()
{
  int n;
  for(;cin>>n,n;)
    {
      long long int s = pre(n);
      if(n>s)
	cout<<"deficient number"<<endl;
      else if(n<s)
	cout<<"abundant number"<<endl;
      else 
	cout<<"perfect number"<<endl;
    }
  
}
### Prompt

Your task is to create 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>
#include<math.h>
using namespace std;

long long int pre(long long int n)
{
  if(n<2)
    return 0;
  long long int c=1;
  for(int i=2;i<=sqrt(n);i++)
    if(n%i==0)
      {
	c+=i;
	if(i!=n/i)
	  c+=n/i;
      }
  return c;
}

int main()
{
  int n;
  for(;cin>>n,n;)
    {
      long long int s = pre(n);
      if(n>s)
	cout<<"deficient number"<<endl;
      else if(n<s)
	cout<<"abundant number"<<endl;
      else 
	cout<<"perfect number"<<endl;
    }
  
}
```