#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;

/* constant */

/* typedef */

/* global variables */

/* subroutines */

int sumdiv(int n) {
  if (n == 1) return 0;
  int sum = 1;
  for (int k = 2; k * k <= n; k++)
    if (n % k == 0) {
      int m = n / k;
      if (k == m) sum += k;
      else sum += (k + m);
    }
  return sum;
}

/* main */

int main() {
  for (;;) {
    int n;
    cin >> n;
    if (n == 0) break;

    int s = sumdiv(n);
    //printf("n=%d, s=%d\n", n, s);
    
    if (n == s) cout << "perfect number" << endl;
    else if (n > s) cout << "deficient number" << endl;
    else cout << "abundant number" << endl;
  }
  
  return 0;
}
### 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<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;

/* constant */

/* typedef */

/* global variables */

/* subroutines */

int sumdiv(int n) {
  if (n == 1) return 0;
  int sum = 1;
  for (int k = 2; k * k <= n; k++)
    if (n % k == 0) {
      int m = n / k;
      if (k == m) sum += k;
      else sum += (k + m);
    }
  return sum;
}

/* main */

int main() {
  for (;;) {
    int n;
    cin >> n;
    if (n == 0) break;

    int s = sumdiv(n);
    //printf("n=%d, s=%d\n", n, s);
    
    if (n == s) cout << "perfect number" << endl;
    else if (n > s) cout << "deficient number" << endl;
    else cout << "abundant number" << endl;
  }
  
  return 0;
}
```