#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
using namespace std;

string solve(int n){
	long long int sum=0;
	for(int i=1; i*i<=n; ++i){
		if(n%i==0&&n!=1){
			sum+=i;
			if(i!=n/i&&i!=1) sum+=n/i;
		}
	}
	//cout << sum << endl;
	if(sum<n) return "deficient number";
	else if(n<sum) return "abundant number";
	else return "perfect number";
}

int main(){
	int N;
	while(cin >> N&&N){
		cout << solve(N) << endl;
	}
}
### Prompt

Your task is to create 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 <algorithm>
#include <vector>
#include <string>
using namespace std;

string solve(int n){
	long long int sum=0;
	for(int i=1; i*i<=n; ++i){
		if(n%i==0&&n!=1){
			sum+=i;
			if(i!=n/i&&i!=1) sum+=n/i;
		}
	}
	//cout << sum << endl;
	if(sum<n) return "deficient number";
	else if(n<sum) return "abundant number";
	else return "perfect number";
}

int main(){
	int N;
	while(cin >> N&&N){
		cout << solve(N) << endl;
	}
}
```