#include <cstdlib>
#include <iostream>
using namespace std;

string calc(int n) {
	int sum = (n == 1 ? 0 : 1);
	for(int i = 2; i * i <= n; ++i) {
		if(n % i == 0) {
			sum += i;
			if(i * i != n) sum += n / i;
		}
	}

	return sum < n ? "deficient number" :
		sum == n ? "perfect number" : "abundant number";
}

int main() {
	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	for(int n; cin >> n && n;) {
		cout << calc(n) << endl;
	}

	return EXIT_SUCCESS;
}
### 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 <cstdlib>
#include <iostream>
using namespace std;

string calc(int n) {
	int sum = (n == 1 ? 0 : 1);
	for(int i = 2; i * i <= n; ++i) {
		if(n % i == 0) {
			sum += i;
			if(i * i != n) sum += n / i;
		}
	}

	return sum < n ? "deficient number" :
		sum == n ? "perfect number" : "abundant number";
}

int main() {
	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	for(int n; cin >> n && n;) {
		cout << calc(n) << endl;
	}

	return EXIT_SUCCESS;
}
```