#include <iostream>
using namespace std;

int main(){
    int n;

    while(cin >> n, n != 0){
        if(n == 1){
            cout << "deficient number" << endl;
            continue;
        }
        int sum = 1;
        for(int i=2; i*i<=n; i++){
            if(n%i == 0) {
                if(n/i != i){
                    sum += (i + n/i);
                }else{
                    sum += i;
                }
            }
        }
        if(sum < n) cout << "deficient number" << endl;
        if(sum == n) cout << "perfect number" << endl;
        if(sum > n) cout << "abundant number" << endl;
    }
    return 0;
}
### Prompt

Please formulate 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
#include <iostream>
using namespace std;

int main(){
    int n;

    while(cin >> n, n != 0){
        if(n == 1){
            cout << "deficient number" << endl;
            continue;
        }
        int sum = 1;
        for(int i=2; i*i<=n; i++){
            if(n%i == 0) {
                if(n/i != i){
                    sum += (i + n/i);
                }else{
                    sum += i;
                }
            }
        }
        if(sum < n) cout << "deficient number" << endl;
        if(sum == n) cout << "perfect number" << endl;
        if(sum > n) cout << "abundant number" << endl;
    }
    return 0;
}
```