#include<iostream>
using namespace std;
bool isPrime(int n){
	if(n==1)return 0;
	if(n==2)return 1;
	int i;
	for(i=3;i*i<=n;i+=2){
		if(n%i==0)return 0;
	}
	return 1;
}
int judge(int n){
	//cout<<"n:"<<n<<endl;
	int i;
	int sum=0;
	for(i=1;i<=n/i;i++){
		if(n%i==0){
			if(i==n);
			else if(i==n/i||i==1)sum+=i;
			else sum+=i+(n/i);
			//cout<<"("<<i<<","<<n/i<<") "<<i*(n/i)<<" "<<sum<<endl;
		}
	}
	if(sum==n)return 0;
	else if(sum<n)return 1;
	else if(sum>n)return 2;
}
int main(){
	int n;

	for(;cin>>n&&n;){
		switch(judge(n)){
			case 0:
				cout<<"perfect number"<<endl;
				break;
			case 1:
				cout<<"deficient number"<<endl;
				break;
			case 2:
				cout<<"abundant number"<<endl;
				break;
		}
	}



	return 0;
}
### Prompt

Your task is to create 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;
bool isPrime(int n){
	if(n==1)return 0;
	if(n==2)return 1;
	int i;
	for(i=3;i*i<=n;i+=2){
		if(n%i==0)return 0;
	}
	return 1;
}
int judge(int n){
	//cout<<"n:"<<n<<endl;
	int i;
	int sum=0;
	for(i=1;i<=n/i;i++){
		if(n%i==0){
			if(i==n);
			else if(i==n/i||i==1)sum+=i;
			else sum+=i+(n/i);
			//cout<<"("<<i<<","<<n/i<<") "<<i*(n/i)<<" "<<sum<<endl;
		}
	}
	if(sum==n)return 0;
	else if(sum<n)return 1;
	else if(sum>n)return 2;
}
int main(){
	int n;

	for(;cin>>n&&n;){
		switch(judge(n)){
			case 0:
				cout<<"perfect number"<<endl;
				break;
			case 1:
				cout<<"deficient number"<<endl;
				break;
			case 2:
				cout<<"abundant number"<<endl;
				break;
		}
	}



	return 0;
}
```