#include <iostream>
#include <algorithm>
#include <cmath>
#define int long long
using namespace std;
signed main() {
	int n;
	while (cin >> n, n) {
		if (n == 1) {
			cout << "deficient number\n";
			continue;
		}
		int sum = 1;
		for (int i = 2; i <= sqrt(n); i++) {
			if (n % i == 0)sum += i + (n / i);
			if (i*i == n)sum -= i;
		}
		if (n == sum)cout << "perfect number\n";
		else cout << (n < sum ? "abundant number\n" : "deficient number\n");
	}
}

### Prompt

Create a solution in cpp for 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>
#include <algorithm>
#include <cmath>
#define int long long
using namespace std;
signed main() {
	int n;
	while (cin >> n, n) {
		if (n == 1) {
			cout << "deficient number\n";
			continue;
		}
		int sum = 1;
		for (int i = 2; i <= sqrt(n); i++) {
			if (n % i == 0)sum += i + (n / i);
			if (i*i == n)sum -= i;
		}
		if (n == sum)cout << "perfect number\n";
		else cout << (n < sum ? "abundant number\n" : "deficient number\n");
	}
}

```