#include <iostream>
#include <math.h>
using namespace std;
void doIt3();

int main(){
doIt3();
}

void doIt3(){
	int n;
	cin >> n;
	while(n != 0){
		if(n == 1){
			cout << "deficient number" << endl;
			cin >> n;
			continue;
		}
		double rt = sqrt(n);
		//まず1の処理
		int sum = 1;
		int i;
		for(i = 2; i < rt; i++){
			if(n % i == 0){
				sum += i;
				sum += n / i;
			}
		}
		if((double)i == rt){
			sum += i;
		}
		if(sum == n){
			cout << "perfect number" << endl;
		}
		else if(sum > n){
			cout << "abundant number" << endl;
		}
		else{
			cout << "deficient number" << endl;
		}
		//cout << sum << endl;
		cin >> n;
	}
	
}
### 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 <math.h>
using namespace std;
void doIt3();

int main(){
doIt3();
}

void doIt3(){
	int n;
	cin >> n;
	while(n != 0){
		if(n == 1){
			cout << "deficient number" << endl;
			cin >> n;
			continue;
		}
		double rt = sqrt(n);
		//まず1の処理
		int sum = 1;
		int i;
		for(i = 2; i < rt; i++){
			if(n % i == 0){
				sum += i;
				sum += n / i;
			}
		}
		if((double)i == rt){
			sum += i;
		}
		if(sum == n){
			cout << "perfect number" << endl;
		}
		else if(sum > n){
			cout << "abundant number" << endl;
		}
		else{
			cout << "deficient number" << endl;
		}
		//cout << sum << endl;
		cin >> n;
	}
	
}
```