#include <cstdio>

#define i64 long long
using namespace std;
int main(){
  i64 r,n;
  while(scanf("%lld",&n),n){
    r=0LL;
    for(i64 i=1LL;i*i<n;i++){
      if(!(n%i)){
        r+=i;if(i>1)r+=n/i;
      }
    }
    // printf("%d %d\n",n,r);
    if(r>n){
      puts("abundant number");
    }
    else if(r==n&&n!=1){
      puts("perfect number");
    }
    else{
      puts("deficient number");
    }
  }
}
### 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 <cstdio>

#define i64 long long
using namespace std;
int main(){
  i64 r,n;
  while(scanf("%lld",&n),n){
    r=0LL;
    for(i64 i=1LL;i*i<n;i++){
      if(!(n%i)){
        r+=i;if(i>1)r+=n/i;
      }
    }
    // printf("%d %d\n",n,r);
    if(r>n){
      puts("abundant number");
    }
    else if(r==n&&n!=1){
      puts("perfect number");
    }
    else{
      puts("deficient number");
    }
  }
}
```