#include <cmath>
#include <cstdio>
#include <iostream>
#include <string>
using namespace std;

int main(){
	while(1){
		long n;
		scanf(" %ld", &n);
		if(n==0) break;
		
		long s=1;
		for(long i=2; i<=sqrt(n); ++i){
			if(n%i==0){
				if(i==n/i) s+=i;
				else s+=i+n/i;
			}
		}
		
		string ans;
		if(n==1 || s<n) ans="deficient number";
		else if(s==n) ans="perfect number";
		else ans="abundant number";
		
		cout << ans << endl;
	}
}
### 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 <cmath>
#include <cstdio>
#include <iostream>
#include <string>
using namespace std;

int main(){
	while(1){
		long n;
		scanf(" %ld", &n);
		if(n==0) break;
		
		long s=1;
		for(long i=2; i<=sqrt(n); ++i){
			if(n%i==0){
				if(i==n/i) s+=i;
				else s+=i+n/i;
			}
		}
		
		string ans;
		if(n==1 || s<n) ans="deficient number";
		else if(s==n) ans="perfect number";
		else ans="abundant number";
		
		cout << ans << endl;
	}
}
```