#include <iostream>
using namespace std;

int main() {
    while (true) {
        int n;
        cin >> n;
        if (n == 0) {
            break;
        }
        int sum = 0;
        for (int i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                if (1 < i && i * i < n) {
                    sum += i + n / i;
                } else if (n != 1) {
                    sum += i;
                }
            }
        }
        if (n > sum) {
            cout << "deficient number" << endl;
        } else if (n < sum) {
            cout << "abundant number" << endl;
        } else {
            cout << "perfect number" << 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>
using namespace std;

int main() {
    while (true) {
        int n;
        cin >> n;
        if (n == 0) {
            break;
        }
        int sum = 0;
        for (int i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                if (1 < i && i * i < n) {
                    sum += i + n / i;
                } else if (n != 1) {
                    sum += i;
                }
            }
        }
        if (n > sum) {
            cout << "deficient number" << endl;
        } else if (n < sum) {
            cout << "abundant number" << endl;
        } else {
            cout << "perfect number" << endl;
        }
    }
}
```