#include<iostream>
#include<string>
using namespace std;
int n;

string hantei(int n) {
	int sum = 0;
	for (int i = 1; i <= n / 4; i++) {
		if (n%i == 0) { sum += i; }
	}
	if (n % 3 == 0) { sum += n / 3; }
	if (n % 2 == 0) { sum += n / 2; }
	if (sum < n) { return "deficient number"; }
	if (sum == n) { return "perfect number"; }
	return "abundant number";
}

int main() {
	while (true) {
		cin >> n;
		if (n == 0) { break; }
		cout << hantei(n) << endl;
	}
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#include<string>
using namespace std;
int n;

string hantei(int n) {
	int sum = 0;
	for (int i = 1; i <= n / 4; i++) {
		if (n%i == 0) { sum += i; }
	}
	if (n % 3 == 0) { sum += n / 3; }
	if (n % 2 == 0) { sum += n / 2; }
	if (sum < n) { return "deficient number"; }
	if (sum == n) { return "perfect number"; }
	return "abundant number";
}

int main() {
	while (true) {
		cin >> n;
		if (n == 0) { break; }
		cout << hantei(n) << endl;
	}
}
```