#include<cstdio>
#include<vector>

#define	mp	make_pair
#define	pb	push_back

using namespace std;

typedef	vector<int>		vi;
typedef	pair<int,int>	pii;

int main(){
	static bool era[10000];
	for(int i=2;i<100;i++){
		if(!era[i])	for(int j=i*i;j<10000;j+=i)	era[j]=true;
	}

	vi pr;
	for(int i=2;i<10000;i++)	if(!era[i])	pr.pb(i);

	for(int n;scanf("%d",&n),n;){
		int nn=n;
		vector<pii> div;
		for(int i=0;i<pr.size()&&n>1;i++){
			if(n%pr[i]==0){
				int m=0;
				do{
					n/=pr[i];
					m++;
				}while(n%pr[i]==0);
				div.pb(mp(pr[i],m));
			}
		}
		if(n>1)	div.pb(mp(n,1));

		long long dsum=1;
		for(int i=0;i<div.size();i++){
			int d=div[i].first,m=div[i].second;
			long long tmp=1;
			for(int j=0;j<m+1;j++)	tmp*=d;
			dsum*=(1-tmp)/(1-d);
		}
		dsum-=nn;

		puts(nn==dsum?"perfect number":nn<dsum?"abundant number":"deficient number");
	}

	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<cstdio>
#include<vector>

#define	mp	make_pair
#define	pb	push_back

using namespace std;

typedef	vector<int>		vi;
typedef	pair<int,int>	pii;

int main(){
	static bool era[10000];
	for(int i=2;i<100;i++){
		if(!era[i])	for(int j=i*i;j<10000;j+=i)	era[j]=true;
	}

	vi pr;
	for(int i=2;i<10000;i++)	if(!era[i])	pr.pb(i);

	for(int n;scanf("%d",&n),n;){
		int nn=n;
		vector<pii> div;
		for(int i=0;i<pr.size()&&n>1;i++){
			if(n%pr[i]==0){
				int m=0;
				do{
					n/=pr[i];
					m++;
				}while(n%pr[i]==0);
				div.pb(mp(pr[i],m));
			}
		}
		if(n>1)	div.pb(mp(n,1));

		long long dsum=1;
		for(int i=0;i<div.size();i++){
			int d=div[i].first,m=div[i].second;
			long long tmp=1;
			for(int j=0;j<m+1;j++)	tmp*=d;
			dsum*=(1-tmp)/(1-d);
		}
		dsum-=nn;

		puts(nn==dsum?"perfect number":nn<dsum?"abundant number":"deficient number");
	}

	return 0;
}
```