#include <bits/stdc++.h>

using namespace std;

int main(void) {
  long long n;
  while (cin >> n, n) {
    long long sum = 0LL;
    for (int i = 1; i * i <= n; i++) {
      if (n % i) continue;
      sum += i;
      if (i * i != n) sum += n / i;
    }
    if (n == 1 || 2 * n > sum) cout << "deficient number\n";
    else if (2 * n < sum) cout << "abundant number\n";
    else cout << "perfect number\n";
  }

  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 <bits/stdc++.h>

using namespace std;

int main(void) {
  long long n;
  while (cin >> n, n) {
    long long sum = 0LL;
    for (int i = 1; i * i <= n; i++) {
      if (n % i) continue;
      sum += i;
      if (i * i != n) sum += n / i;
    }
    if (n == 1 || 2 * n > sum) cout << "deficient number\n";
    else if (2 * n < sum) cout << "abundant number\n";
    else cout << "perfect number\n";
  }

  return 0;
}

```