//42-47
#include<iostream>

using namespace std;

int main(){
  bool cmp[10001]={true,true};
  for(int i=2;i<=10000;i++){
    if(!cmp[i]){
      for(int j=2;i*j<=10000;j++){
	cmp[i*j]=true;
      }
    }
  }
  for(int n;cin>>n,n;){
    int nt=n;
    int ans=1;
    for(int i=0;i<=10001;i++){
      if(!cmp[i]){
	int f=1;
	int s=1;
	while(nt%i==0){
	  nt/=i;
	  f*=i;
	  s+=f;
	}
	ans*=s;
      }
    }
    if(nt!=1){
      ans*=(1+nt);
    }
    ans-=n;
    if(n>ans){
      cout<<"deficient number"<<endl;
    }else if(n<ans){
      cout<<"abundant number"<<endl;
    }else{
      cout<<"perfect number"<<endl;
    }
  }
  return 0;
}
### 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
//42-47
#include<iostream>

using namespace std;

int main(){
  bool cmp[10001]={true,true};
  for(int i=2;i<=10000;i++){
    if(!cmp[i]){
      for(int j=2;i*j<=10000;j++){
	cmp[i*j]=true;
      }
    }
  }
  for(int n;cin>>n,n;){
    int nt=n;
    int ans=1;
    for(int i=0;i<=10001;i++){
      if(!cmp[i]){
	int f=1;
	int s=1;
	while(nt%i==0){
	  nt/=i;
	  f*=i;
	  s+=f;
	}
	ans*=s;
      }
    }
    if(nt!=1){
      ans*=(1+nt);
    }
    ans-=n;
    if(n>ans){
      cout<<"deficient number"<<endl;
    }else if(n<ans){
      cout<<"abundant number"<<endl;
    }else{
      cout<<"perfect number"<<endl;
    }
  }
  return 0;
}
```