#include<stdio.h>
#include<iostream>
using namespace std;
int main(){

int n;
while(1){

scanf("%d",&n);if(n==0)break;
int s=0;int t=0;
for(int i=1;i*i<=n;i++)
{t=n/i;

	if(n%i==0){s+=i+t;
	if(n==t||t==i)s-=t;
	
	}
//cout<<t<<endl;
}
if(n==1)s=0;
//cout<<s<<endl;
if(s>n)printf("abundant number\n");
if(s==n)printf("perfect number\n");
if(s<n)printf("deficient number\n");
}
return 0;
}
### Prompt

Your challenge is to write 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>
#include<iostream>
using namespace std;
int main(){

int n;
while(1){

scanf("%d",&n);if(n==0)break;
int s=0;int t=0;
for(int i=1;i*i<=n;i++)
{t=n/i;

	if(n%i==0){s+=i+t;
	if(n==t||t==i)s-=t;
	
	}
//cout<<t<<endl;
}
if(n==1)s=0;
//cout<<s<<endl;
if(s>n)printf("abundant number\n");
if(s==n)printf("perfect number\n");
if(s<n)printf("deficient number\n");
}
return 0;
}
```