#include<bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
  int n;
  while(cin>>n,n){
    int ans=1;
    int m=n;
    for(int i=2;i*i<=n;i++){
      if(m%i) continue;
      int tmp=1,res=1;
      while(m%i==0){
	m/=i;
	res*=i;
	tmp+=res;
      }
      ans*=tmp;
      if(ans>n*2) break;
    }
    if(m!=1) ans*=m+1;
    if(ans==n*2) cout<<"perfect number"<<endl;
    if(ans<n*2) cout<<"deficient number"<<endl;
    if(ans>n*2) cout<<"abundant number"<<endl;
  }
  return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;
#define int long long
signed main(){
  int n;
  while(cin>>n,n){
    int ans=1;
    int m=n;
    for(int i=2;i*i<=n;i++){
      if(m%i) continue;
      int tmp=1,res=1;
      while(m%i==0){
	m/=i;
	res*=i;
	tmp+=res;
      }
      ans*=tmp;
      if(ans>n*2) break;
    }
    if(m!=1) ans*=m+1;
    if(ans==n*2) cout<<"perfect number"<<endl;
    if(ans<n*2) cout<<"deficient number"<<endl;
    if(ans>n*2) cout<<"abundant number"<<endl;
  }
  return 0;
}
```