#include<bits/stdc++.h>
using namespace std;
int main(){

  int n,ans,i,ika;
  
  scanf("%d",&n);
  while(n != 0){
    
    ika = n;
    i = 2;
    if(n == 1)ans = 0;
    else ans = 1;
    
    while(ika > i){
      
      if(n%i == 0){
	ans += i;
	if(i*i != n) ans += n/i;
	ika = n/i;
      }
      
      i++;
    }
    
    if(ans == n) printf("perfect number\n");
    else if(ans < n) printf("deficient number\n");
    else if(ans > n) printf("abundant number\n");
    scanf("%d",&n);
  }
  
  return (0);
}
 
### Prompt

Generate 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<bits/stdc++.h>
using namespace std;
int main(){

  int n,ans,i,ika;
  
  scanf("%d",&n);
  while(n != 0){
    
    ika = n;
    i = 2;
    if(n == 1)ans = 0;
    else ans = 1;
    
    while(ika > i){
      
      if(n%i == 0){
	ans += i;
	if(i*i != n) ans += n/i;
	ika = n/i;
      }
      
      i++;
    }
    
    if(ans == n) printf("perfect number\n");
    else if(ans < n) printf("deficient number\n");
    else if(ans > n) printf("abundant number\n");
    scanf("%d",&n);
  }
  
  return (0);
}
 
```