#include<stdio.h>
int main(void){
	int n;
	while(1){
		int i,s=0,squareroot;
		scanf("%d",&n);
		if(n==0)break;
		
		for(i=1;;i++){
			if(i*i<=n&&n<(i+1)*(i+1)){
				squareroot=i;
				break;
			}
			
		}
		for(i=1;i<=squareroot;i++){
			if(n%i==0){
				s+=n/i;
				s+=i;
				if(i==n/i)s-=i;
			}
			
			
			
			
		}
		s-=n;
		
		if(n==s)printf("perfect number\n");
		else if(n>s)printf("deficient number\n");
		else printf("abundant number\n");
		
	}
	return 0;
}
### Prompt

Please formulate 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<stdio.h>
int main(void){
	int n;
	while(1){
		int i,s=0,squareroot;
		scanf("%d",&n);
		if(n==0)break;
		
		for(i=1;;i++){
			if(i*i<=n&&n<(i+1)*(i+1)){
				squareroot=i;
				break;
			}
			
		}
		for(i=1;i<=squareroot;i++){
			if(n%i==0){
				s+=n/i;
				s+=i;
				if(i==n/i)s-=i;
			}
			
			
			
			
		}
		s-=n;
		
		if(n==s)printf("perfect number\n");
		else if(n>s)printf("deficient number\n");
		else printf("abundant number\n");
		
	}
	return 0;
}
```