#include <iostream>
#include <stdio.h>
#include <math.h>

using namespace std;

int main(){
  int n,sum,cnt;
  while (true){
    cin>>n;
    if (n==0) return 0;
    sum=cnt=0;
    for (int i=1;i<=sqrt(n);i++)
      if (n%i==0)
        sum+=i+n/i;
    if (sqrt(n)==(int)sqrt(n))
      sum-=n+(int)sqrt(n);
    else
      sum-=n;
    if (sum==n)
      puts("perfect number");
    else if (sum<n)
      puts("deficient number");
    else
      puts("abundant number");
  }
}
### 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>
#include <stdio.h>
#include <math.h>

using namespace std;

int main(){
  int n,sum,cnt;
  while (true){
    cin>>n;
    if (n==0) return 0;
    sum=cnt=0;
    for (int i=1;i<=sqrt(n);i++)
      if (n%i==0)
        sum+=i+n/i;
    if (sqrt(n)==(int)sqrt(n))
      sum-=n+(int)sqrt(n);
    else
      sum-=n;
    if (sum==n)
      puts("perfect number");
    else if (sum<n)
      puts("deficient number");
    else
      puts("abundant number");
  }
}
```