#include <iostream>
using namespace std;

#define MAX 100000001
bool p[MAX];

int main()
{
	int N;
	/*
	for (int i=0; i<MAX; i++)
		p[i] = true;

	p[0] = [1] = false;
	for (int i=2; i*i<MAX; i++)
	{
		if (p[i])
		{
			for (int k=i; k<MAX; k+=i)
				p[k] = false;
		}
	}*/

	while (cin >> N, N)
	{
		int sum = 1;
		if (N == 1) sum = 0;
		for (int i=2; i*i<=N; i++)
		{
			if (N%i == 0)
			{
				sum += i;
				sum += N/i;
			
				if (i == N/i)
				{
					sum -= i;
				}
			}
		}

		if (sum < N) cout << "deficient number" << endl;
		else if (sum == N) cout << "perfect number" << endl;
		else cout << "abundant number" << endl;
	}

	return 0;
}
### Prompt

Please create a solution in Cpp 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;

#define MAX 100000001
bool p[MAX];

int main()
{
	int N;
	/*
	for (int i=0; i<MAX; i++)
		p[i] = true;

	p[0] = [1] = false;
	for (int i=2; i*i<MAX; i++)
	{
		if (p[i])
		{
			for (int k=i; k<MAX; k+=i)
				p[k] = false;
		}
	}*/

	while (cin >> N, N)
	{
		int sum = 1;
		if (N == 1) sum = 0;
		for (int i=2; i*i<=N; i++)
		{
			if (N%i == 0)
			{
				sum += i;
				sum += N/i;
			
				if (i == N/i)
				{
					sum -= i;
				}
			}
		}

		if (sum < N) cout << "deficient number" << endl;
		else if (sum == N) cout << "perfect number" << endl;
		else cout << "abundant number" << endl;
	}

	return 0;
}
```