#include<iostream>
#include<cmath>
#include<vector>
using namespace std;

int main(){
  int n;
  while(1){
    cin >> n;
    if(!n)break;

    vector<int> d;
    int r = (int)sqrt(n);
    for(int i=2;i<=r;i++){
      if( !(n%i) ){
	d.push_back(i);
	if(i!=n/i)d.push_back(n/i);
      }
    }

    int sum = 0;
    if(n!=1){
      sum += 1;
      for(int i=0;i<(int)d.size();i++){
	sum += d[i];
      }
    }
    if(sum<n)cout << "deficient number" << endl;
    else if(sum>n)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<cmath>
#include<vector>
using namespace std;

int main(){
  int n;
  while(1){
    cin >> n;
    if(!n)break;

    vector<int> d;
    int r = (int)sqrt(n);
    for(int i=2;i<=r;i++){
      if( !(n%i) ){
	d.push_back(i);
	if(i!=n/i)d.push_back(n/i);
      }
    }

    int sum = 0;
    if(n!=1){
      sum += 1;
      for(int i=0;i<(int)d.size();i++){
	sum += d[i];
      }
    }
    if(sum<n)cout << "deficient number" << endl;
    else if(sum>n)cout << "abundant number" << endl;
    else cout << "perfect number" << endl;
  }
}
```