#include <iostream>

using namespace std;
int perfect_number[5] = {6, 28, 496, 8128, 33550336};

void solve()
{
	int N;
	while(cin >> N, N)
	{
		bool perfect_flag = false;
		for(int i = 0; i < 5; ++i)
		{
			if(perfect_number[i] == N)
			{
				cout << "perfect number" << endl;
				perfect_flag = true;
				break;
			}
		}
		if(perfect_flag)
		{
			continue;
		}
		int sum = -N;
		for(int i = 1; i * i <= N; ++i)
		{
			if(N % i == 0)
			{
				sum += i + N / i;
				if(i == N / i)
				{
					sum -= i;
				}
			}
		}
		if(sum < N)
		{
			cout << "deficient number" << endl;
		}
		else
		{
			cout << "abundant number" << endl;
		}
	}
}

int main()
{
	solve();
	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 <iostream>

using namespace std;
int perfect_number[5] = {6, 28, 496, 8128, 33550336};

void solve()
{
	int N;
	while(cin >> N, N)
	{
		bool perfect_flag = false;
		for(int i = 0; i < 5; ++i)
		{
			if(perfect_number[i] == N)
			{
				cout << "perfect number" << endl;
				perfect_flag = true;
				break;
			}
		}
		if(perfect_flag)
		{
			continue;
		}
		int sum = -N;
		for(int i = 1; i * i <= N; ++i)
		{
			if(N % i == 0)
			{
				sum += i + N / i;
				if(i == N / i)
				{
					sum -= i;
				}
			}
		}
		if(sum < N)
		{
			cout << "deficient number" << endl;
		}
		else
		{
			cout << "abundant number" << endl;
		}
	}
}

int main()
{
	solve();
	return(0);
}
```