#include<bits/stdc++.h>
#define range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) range(i,0,b)
#define pb(a) push_back(a)
#define all(a) (a).begin(), (a).end()
#define debug(x) cout << "debug " << x << endl;
using namespace std;

int main(){
    int n;
    while(cin >> n,n){
        int sum = 0;
        for(int i = 1; i * i <= n; i++){
            if(n % i == 0 && i != n){
                sum+=i;
                if(n / i != i && n / i != n) sum+=n/i;
            }
        }
        if(sum > n){
            cout << "abundant number" << endl;
        }else if(sum < n){
            cout << "deficient number" << endl;
        }else{
            cout << "perfect number" << endl;
        }
    }
}
### 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<bits/stdc++.h>
#define range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) range(i,0,b)
#define pb(a) push_back(a)
#define all(a) (a).begin(), (a).end()
#define debug(x) cout << "debug " << x << endl;
using namespace std;

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