#include<bits/stdc++.h>
using namespace std;

int main(){
  int n;
  while(cin>>n,n){
    int a=2;
    int tmp=n;
    vector<int> ans;
    ans.push_back(1);
    while((int)sqrt(n)>=a && n!=1){
      if(n%a==0){
	ans.push_back(a);
	if(a!=n/a)
	  ans.push_back(n/a);
      }
      a++;
    }
    int b=0;
    for(int i=0;i<ans.size();i++){
      if(ans[i]==tmp)continue;
      b+=ans[i];
    }
    if(tmp==b && tmp != 1){
      cout << "perfect number" << endl;
    }else if(tmp > b){
      cout << "deficient number" << endl;
    }else{
      cout << "abundant number" << endl;
    }
  }
  return 0;
}
### Prompt

Please formulate 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<bits/stdc++.h>
using namespace std;

int main(){
  int n;
  while(cin>>n,n){
    int a=2;
    int tmp=n;
    vector<int> ans;
    ans.push_back(1);
    while((int)sqrt(n)>=a && n!=1){
      if(n%a==0){
	ans.push_back(a);
	if(a!=n/a)
	  ans.push_back(n/a);
      }
      a++;
    }
    int b=0;
    for(int i=0;i<ans.size();i++){
      if(ans[i]==tmp)continue;
      b+=ans[i];
    }
    if(tmp==b && tmp != 1){
      cout << "perfect number" << endl;
    }else if(tmp > b){
      cout << "deficient number" << endl;
    }else{
      cout << "abundant number" << endl;
    }
  }
  return 0;
}
```