#include<iostream>
#include<string>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
int main(void){

	int n;
	int s=0;

	for(;;){
	    s=0;
        cin>>n;
        int x=(int)sqrt(n);
        if(n==0)break;
        for(int i=2;i<=x;i++){
            if(n%i==0)s+=i,s+=n/i;//,cout<<"-"<<i<<endl;
        }
    if(x*x==n)s-=x;
    s++;
    if(n<=4)s=0;

	//cout<<s<<" "<<x<<endl;
	if(s==n)cout<<"perfect number"<<endl;
	else if(s>n)cout<<"abundant number"<<endl;
	else if(s<n)cout<<"deficient number"<<endl;

	}


	return 0;
}
### Prompt

Generate 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>
#include<string>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
int main(void){

	int n;
	int s=0;

	for(;;){
	    s=0;
        cin>>n;
        int x=(int)sqrt(n);
        if(n==0)break;
        for(int i=2;i<=x;i++){
            if(n%i==0)s+=i,s+=n/i;//,cout<<"-"<<i<<endl;
        }
    if(x*x==n)s-=x;
    s++;
    if(n<=4)s=0;

	//cout<<s<<" "<<x<<endl;
	if(s==n)cout<<"perfect number"<<endl;
	else if(s>n)cout<<"abundant number"<<endl;
	else if(s<n)cout<<"deficient number"<<endl;

	}


	return 0;
}
```