#include <iostream>
#include <string>
#include <algorithm>
#include <set>
#include <vector>
#include <cstring>
#include <climits>
#include <queue>
#include <map>
#include <sstream>
using namespace std;

#define REP(i,a,n) for(int i=(a); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define EPS 1e-9

int main(){
	long long n;
	while(scanf("%lld",&n),n){
		long long sum = 1;
		long long K = (long long)n;
		for(long long i=2; i*i<=n; i++){
			if( n%i==0 ){
				long long cnt = 1;
				for(long long k=(long long)i;n%i==0;n/=i,k*=i)cnt+=k;
				sum *= (long long)cnt;
				//printf("i:%lld , cnt:%lld  ,  sum:%lld\n",i,cnt,sum);
			}
			//if( sum-2*K>0 ){ break; }
		}
		if( n>1 )sum*=(long long)(n+1);
		//printf("n:%lld , sum:%lld\n",n,sum);
		if( sum-2*K==0 ){
			puts("perfect number");
		}else if( sum-2*K<0 ){
			puts("deficient number");
		}else{
			puts("abundant number");
		}
	}
	return 0;
}
### 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 <set>
#include <vector>
#include <cstring>
#include <climits>
#include <queue>
#include <map>
#include <sstream>
using namespace std;

#define REP(i,a,n) for(int i=(a); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define EPS 1e-9

int main(){
	long long n;
	while(scanf("%lld",&n),n){
		long long sum = 1;
		long long K = (long long)n;
		for(long long i=2; i*i<=n; i++){
			if( n%i==0 ){
				long long cnt = 1;
				for(long long k=(long long)i;n%i==0;n/=i,k*=i)cnt+=k;
				sum *= (long long)cnt;
				//printf("i:%lld , cnt:%lld  ,  sum:%lld\n",i,cnt,sum);
			}
			//if( sum-2*K>0 ){ break; }
		}
		if( n>1 )sum*=(long long)(n+1);
		//printf("n:%lld , sum:%lld\n",n,sum);
		if( sum-2*K==0 ){
			puts("perfect number");
		}else if( sum-2*K<0 ){
			puts("deficient number");
		}else{
			puts("abundant number");
		}
	}
	return 0;
}
```