#include <bits/stdc++.h>
#define LL long long
using namespace std;
int main(){
    LL n;
    vector<LL> vc;
    while(cin>>n){
        LL temp=n,res=0;
        if(n==0){return 0;}
        for(int i=1; i*i<=n; i++){
            if(n%i==0){
                if(0==n%i){
                    res+=i;
                    if(i!=n/i){res+=n/i;}
                }
            }
        }
        res-=temp;
        cout << (res==temp?"perfect number":res<temp?"deficient number":"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>
#define LL long long
using namespace std;
int main(){
    LL n;
    vector<LL> vc;
    while(cin>>n){
        LL temp=n,res=0;
        if(n==0){return 0;}
        for(int i=1; i*i<=n; i++){
            if(n%i==0){
                if(0==n%i){
                    res+=i;
                    if(i!=n/i){res+=n/i;}
                }
            }
        }
        res-=temp;
        cout << (res==temp?"perfect number":res<temp?"deficient number":"abundant number") << endl;
    }    
    return 0;
}
```