#include <stdio.h>
#pragma warning(disable : 4996)
int n;
int main() {
	while (true) {
		scanf("%d", &n);
		if (n == 0) break;
		int ret = n > 1 ? 1 : 0;
		for (int i = 2; i * i <= n; i++) {
			if (n % i == 0) {
				if (i * i != n) {
					ret += i + n / i;
				}
				else {
					ret += i;
				}
			}
		}
		if (n > ret) {
			printf("deficient number\n");
		}
		else if (n == ret) {
			printf("perfect number\n");
		}
		else {
			printf("abundant number\n");
		}
	}
	return 0;
}
### Prompt

In CPP, your task is to solve 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>
#pragma warning(disable : 4996)
int n;
int main() {
	while (true) {
		scanf("%d", &n);
		if (n == 0) break;
		int ret = n > 1 ? 1 : 0;
		for (int i = 2; i * i <= n; i++) {
			if (n % i == 0) {
				if (i * i != n) {
					ret += i + n / i;
				}
				else {
					ret += i;
				}
			}
		}
		if (n > ret) {
			printf("deficient number\n");
		}
		else if (n == ret) {
			printf("perfect number\n");
		}
		else {
			printf("abundant number\n");
		}
	}
	return 0;
}
```