#include<cmath>
#include<cctype>
#include<cstdlib>
#include<cstdio>
#include<climits>
#include<iostream>
#include<algorithm>
#include<vector>
#include<deque>
#include<string>
#include<queue>
#include<map>
#include<set>
#include<utility>
#include<queue>

using namespace std;

void solve(int n){
  int s = 0;
  for(int i = 1;i*i <= n;i++){
    if(n % i == 0){
      s += i + (n/i);
    }
    if(i*i == n)s -= i;
  }
  s -= n;
  if(n == 1) cout << "deficient number";
  else if(s == n)cout << "perfect number";
  else if(s < n)cout << "deficient number";
  else if(s > n)cout << "abundant number";
  cout << endl;
} 

int main(){
  ios::sync_with_stdio(false);
  int n = 0;
  while(cin >> n && n != 0){
    solve(n);
  }
  return 0;
}
### Prompt

Please create a solution in CPP 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
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<cstdio>
#include<climits>
#include<iostream>
#include<algorithm>
#include<vector>
#include<deque>
#include<string>
#include<queue>
#include<map>
#include<set>
#include<utility>
#include<queue>

using namespace std;

void solve(int n){
  int s = 0;
  for(int i = 1;i*i <= n;i++){
    if(n % i == 0){
      s += i + (n/i);
    }
    if(i*i == n)s -= i;
  }
  s -= n;
  if(n == 1) cout << "deficient number";
  else if(s == n)cout << "perfect number";
  else if(s < n)cout << "deficient number";
  else if(s > n)cout << "abundant number";
  cout << endl;
} 

int main(){
  ios::sync_with_stdio(false);
  int n = 0;
  while(cin >> n && n != 0){
    solve(n);
  }
  return 0;
}
```