#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<climits>
#include<cmath>
using namespace std;
#define MAX 100000010

int n;
void init(){
  n=0;
}

bool input(){
  cin>>n;
  return n;
}

string solve(){
  int ans=1;
  for(int i=2;i*i<n;i++){
    if(n%i==0){
      ans+=i+n/i;
    }
  }

  if(n==1)ans=0;
  if(ans==n){
    return "perfect number";
  }else if(ans<n){
    return "deficient number";
  }else if(ans>n){
    return "abundant number";
  }
}


int main(){
  while(init(),input()){
    cout<<solve()<<endl;
  }
}
### Prompt

Generate 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<cstdio>
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<climits>
#include<cmath>
using namespace std;
#define MAX 100000010

int n;
void init(){
  n=0;
}

bool input(){
  cin>>n;
  return n;
}

string solve(){
  int ans=1;
  for(int i=2;i*i<n;i++){
    if(n%i==0){
      ans+=i+n/i;
    }
  }

  if(n==1)ans=0;
  if(ans==n){
    return "perfect number";
  }else if(ans<n){
    return "deficient number";
  }else if(ans>n){
    return "abundant number";
  }
}


int main(){
  while(init(),input()){
    cout<<solve()<<endl;
  }
}
```