#include <iostream>

using namespace std;

void per(long long int n){
	long long int res=0;
	if(n>1)res++;
	for(int a=2;a*a<=n;a++){
		if(n%a==0){
			if(a==n);
			else{
				if(a==n/a)res+=a;
				else res+=a+n/a;
			}
		}
	}
	if(res>n)cout << "abundant number" << endl;
	if(res==n)cout << "perfect number" << endl;
	if(res<n)cout << "deficient number" << endl;
}

int main(void){
	long long int n;
	while(1){
		cin >> n;
		if(n==0)break;
		per(n);
	}
    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;

void per(long long int n){
	long long int res=0;
	if(n>1)res++;
	for(int a=2;a*a<=n;a++){
		if(n%a==0){
			if(a==n);
			else{
				if(a==n/a)res+=a;
				else res+=a+n/a;
			}
		}
	}
	if(res>n)cout << "abundant number" << endl;
	if(res==n)cout << "perfect number" << endl;
	if(res<n)cout << "deficient number" << endl;
}

int main(void){
	long long int n;
	while(1){
		cin >> n;
		if(n==0)break;
		per(n);
	}
    return 0;
}
```