#include <stdio.h>
#include <math.h>

int main(void){
	int i, n, sum;
	while(1){
		scanf("%d",&n);
		if(!n)
			break;
		sum = 0;
		for(i = 1;i <= sqrt((double)n);i++){
			if(n % i == 0 && n != 1){
				sum += i;
				if(i != 1 && i != n / i)
					sum += n / i;
			}
		}
		if(sum == n)
			printf("perfect number\n");
		else if(sum < n)
			printf("deficient number\n");
		else
			printf("abundant number\n");
	}
	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 <stdio.h>
#include <math.h>

int main(void){
	int i, n, sum;
	while(1){
		scanf("%d",&n);
		if(!n)
			break;
		sum = 0;
		for(i = 1;i <= sqrt((double)n);i++){
			if(n % i == 0 && n != 1){
				sum += i;
				if(i != 1 && i != n / i)
					sum += n / i;
			}
		}
		if(sum == n)
			printf("perfect number\n");
		else if(sum < n)
			printf("deficient number\n");
		else
			printf("abundant number\n");
	}
	return 0;
}
```