#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int main(){
	while(1){
        int n;
        cin >> n;
        if(n == 0) break;
        int sum = 0;
        for(int i = 1; i * i <= n; i++){
            if(n % i == 0 && n != 1){
                sum += i;
                if(n/i != n)sum += n / i;
                if(i * i == n)sum -= i;
            }
        }
        if(sum < n) cout << "deficient number" << endl;
        else if(sum > n) cout << "abundant number" << endl;
        else if(sum == n) cout << "perfect number" << endl;
	}
	return 0;
}
### Prompt

Please provide a Cpp coded solution 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<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int main(){
	while(1){
        int n;
        cin >> n;
        if(n == 0) break;
        int sum = 0;
        for(int i = 1; i * i <= n; i++){
            if(n % i == 0 && n != 1){
                sum += i;
                if(n/i != n)sum += n / i;
                if(i * i == n)sum -= i;
            }
        }
        if(sum < n) cout << "deficient number" << endl;
        else if(sum > n) cout << "abundant number" << endl;
        else if(sum == n) cout << "perfect number" << endl;
	}
	return 0;
}
```