// AOJ 2101 (http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2101)
#include<cstdio>

int divS( int n )
{
	int ret = 0;

	for( int i = 1; i*i <= n; ++i ) if( n % i == 0 )
	{
		ret += i;

		if( i != n/i )
			ret += n/i;
	}

	return ret-n;
}

int main()
{
	int N;
	while( scanf( "%d", &N ), N )
	{
		int S = divS(N);

		puts( N==S?"perfect number":N>S?"deficient number":"abundant number" );
	}

	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
// AOJ 2101 (http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2101)
#include<cstdio>

int divS( int n )
{
	int ret = 0;

	for( int i = 1; i*i <= n; ++i ) if( n % i == 0 )
	{
		ret += i;

		if( i != n/i )
			ret += n/i;
	}

	return ret-n;
}

int main()
{
	int N;
	while( scanf( "%d", &N ), N )
	{
		int S = divS(N);

		puts( N==S?"perfect number":N>S?"deficient number":"abundant number" );
	}

	return 0;
}
```