// Undone
#include <bits/stdc++.h>
using namespace std;
signed main() {
  int N;
  while (cin >> N, N) {
    int ans = (N == 1) ? 0 : 1;
    for (int i = 2; i*i < N; i++) {
      if (N % i == 0) {
        ans += i + N/i;
      }
      if (ans > N) break;
    }
    if (ans == N) {
      cout << "perfect number" << endl;
    } else if (ans > N) {
      cout << "abundant number" << endl;
    } else if (ans < N) {
      cout << "deficient number" << endl;
    }
  }
}



### Prompt

Your task is to create 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
// Undone
#include <bits/stdc++.h>
using namespace std;
signed main() {
  int N;
  while (cin >> N, N) {
    int ans = (N == 1) ? 0 : 1;
    for (int i = 2; i*i < N; i++) {
      if (N % i == 0) {
        ans += i + N/i;
      }
      if (ans > N) break;
    }
    if (ans == N) {
      cout << "perfect number" << endl;
    } else if (ans > N) {
      cout << "abundant number" << endl;
    } else if (ans < N) {
      cout << "deficient number" << endl;
    }
  }
}



```