#include <iostream>
#include <vector>
#include <math.h>

int main(void){

	int N, S;
	int sq_N;

	while(true){
		std::cin >> N;
		if(N==0) break;

		S = 0;
		sq_N = (int)sqrt(N);

		for(int i=1; i<=sq_N; i++){
			if(i==N) break;
			if(N%i==0){
				S+=i;
				if(N!=N/i && i!=N/i) S+=N/i;
			}
		}

		if(N==S) std::cout << "perfect number" << std::endl;
		else if(N>S) std::cout << "deficient number" << std::endl;
		else if(N<S) std::cout << "abundant number" << std::endl;
	
	}
	return 0;
}


### Prompt

Develop a solution in Cpp to the problem described below:

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 <vector>
#include <math.h>

int main(void){

	int N, S;
	int sq_N;

	while(true){
		std::cin >> N;
		if(N==0) break;

		S = 0;
		sq_N = (int)sqrt(N);

		for(int i=1; i<=sq_N; i++){
			if(i==N) break;
			if(N%i==0){
				S+=i;
				if(N!=N/i && i!=N/i) S+=N/i;
			}
		}

		if(N==S) std::cout << "perfect number" << std::endl;
		else if(N>S) std::cout << "deficient number" << std::endl;
		else if(N<S) std::cout << "abundant number" << std::endl;
	
	}
	return 0;
}


```