#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
typedef long long int ll;
#define BIG_NUM 2000000000
using namespace std;

int main(){

	ll N,sum,limit,tmp,calc,count;

	while(true){
		scanf("%lld",&N);
		if(N == 0)break;

		if(N <= 5){
			printf("deficient number\n");
			continue;
		}

		sum = 1;
		limit = sqrt(N);
		tmp = N;

		calc = 1;
		count = 1;
		while(tmp%2 == 0){
			tmp /= 2;
			calc += pow(2,count);
			count++;
		}

		sum *= calc;

		for(int i = 3; i <= limit; i+=2){
			calc = 1;
			count = 1;
			while(tmp%i == 0){
				tmp /= i;
				calc += pow(i,count);
				count++;
			}
			sum *= calc;
		}
		if(tmp < N && tmp != 1){
			sum *= (1+tmp);
		}

		sum -= N;

		if(sum == N){
			printf("perfect number\n");
		}else if(sum > N){
			printf("abundant number\n");
		}else{
			printf("deficient number\n");
		}
	}
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <stdio.h>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
typedef long long int ll;
#define BIG_NUM 2000000000
using namespace std;

int main(){

	ll N,sum,limit,tmp,calc,count;

	while(true){
		scanf("%lld",&N);
		if(N == 0)break;

		if(N <= 5){
			printf("deficient number\n");
			continue;
		}

		sum = 1;
		limit = sqrt(N);
		tmp = N;

		calc = 1;
		count = 1;
		while(tmp%2 == 0){
			tmp /= 2;
			calc += pow(2,count);
			count++;
		}

		sum *= calc;

		for(int i = 3; i <= limit; i+=2){
			calc = 1;
			count = 1;
			while(tmp%i == 0){
				tmp /= i;
				calc += pow(i,count);
				count++;
			}
			sum *= calc;
		}
		if(tmp < N && tmp != 1){
			sum *= (1+tmp);
		}

		sum -= N;

		if(sum == N){
			printf("perfect number\n");
		}else if(sum > N){
			printf("abundant number\n");
		}else{
			printf("deficient number\n");
		}
	}
	return 0;
}
```