#include<iostream>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<utility>
#include<vector>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<time.h>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define mp make_pair
#define all(in) in.begin(),in.end()
const double PI=acos(-1);
const double EPS=1e-10;
const int inf=1e8;
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
int main(){
	int n;
	while(cin>>n,n){
		int sum=0;		
		double d=sqrt(n);
		loop(i,2,1e9){
			if(i>d)break;
			if(n%i==0){
				if(i!=n/i)sum+=i+n/i;
				else sum+=i;
			}
			if(n<sum)break;
		}
		if(n!=1)sum++;
		//cout<<n<<" "<<sum<<endl;
		if(sum==n)cout<<"perfect number"<<endl;
		else if(n<sum)cout<<"abundant number"<<endl;
		else cout<<"deficient 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<iostream>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<utility>
#include<vector>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<time.h>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define mp make_pair
#define all(in) in.begin(),in.end()
const double PI=acos(-1);
const double EPS=1e-10;
const int inf=1e8;
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
int main(){
	int n;
	while(cin>>n,n){
		int sum=0;		
		double d=sqrt(n);
		loop(i,2,1e9){
			if(i>d)break;
			if(n%i==0){
				if(i!=n/i)sum+=i+n/i;
				else sum+=i;
			}
			if(n<sum)break;
		}
		if(n!=1)sum++;
		//cout<<n<<" "<<sum<<endl;
		if(sum==n)cout<<"perfect number"<<endl;
		else if(n<sum)cout<<"abundant number"<<endl;
		else cout<<"deficient number"<<endl;
	}
}
```