#include<cstdio>
void solve(int n){
	if(n==1){printf("deficient number\n"); return;}
	int sum=1;
	for(int i=2;n>=i*i;i++){
		if(n%i==0&&n!=i*i)
 	 	sum+=i+(n/i);
     	if(sum>n){printf("abundant number\n"); return;}
 	}
	if(sum==n)printf("perfect number\n");
	else printf("deficient number\n");
}

int main(){
	int n,ans;
	while(scanf("%d",&n)){
	if(n==0)break;
	solve(n);
	}
  return 0;	
}
### Prompt

In cpp, your task is to solve 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>
void solve(int n){
	if(n==1){printf("deficient number\n"); return;}
	int sum=1;
	for(int i=2;n>=i*i;i++){
		if(n%i==0&&n!=i*i)
 	 	sum+=i+(n/i);
     	if(sum>n){printf("abundant number\n"); return;}
 	}
	if(sum==n)printf("perfect number\n");
	else printf("deficient number\n");
}

int main(){
	int n,ans;
	while(scanf("%d",&n)){
	if(n==0)break;
	solve(n);
	}
  return 0;	
}
```