#include <cstdio>
#include <math.h>
#include <iostream>
using namespace std;

int solve(int n)
{
	int total = 0;
	for (int i = 1; i * i <= n ; i++)
	{
		if (n % i == 0)
		{
			total += i;
			if (i * i != n) total += n / i;
		}
	}
	total -= n;
	return total - n;
}

int main()
{
	int n;
	while (scanf("%d", &n) != EOF)
	{
		if (n == 0) return 0;
		int result = solve(n);
		if (result == 0)
			puts("perfect number");
		else if (result < 0)
			puts("deficient number");
		else if (result > 0)
			puts("abundant number");
	}
}
### 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 <cstdio>
#include <math.h>
#include <iostream>
using namespace std;

int solve(int n)
{
	int total = 0;
	for (int i = 1; i * i <= n ; i++)
	{
		if (n % i == 0)
		{
			total += i;
			if (i * i != n) total += n / i;
		}
	}
	total -= n;
	return total - n;
}

int main()
{
	int n;
	while (scanf("%d", &n) != EOF)
	{
		if (n == 0) return 0;
		int result = solve(n);
		if (result == 0)
			puts("perfect number");
		else if (result < 0)
			puts("deficient number");
		else if (result > 0)
			puts("abundant number");
	}
}
```