#include<cstdio>
#include<cmath>
int yaksu(int a){
        if (a==1) return 1;
        int res=a;
        for(int i=sqrt(a);i>0;i--){
                if(a%i==0){
                        res-=i;
                        if(i!=a/i&&i!=1) res=res-a/i;
                }
                if(res<0) break;
        }
        return res;
}

int main(){
        int a,tk;
        while(1){
        scanf("%d",&a);
        if(a==0) return 0;
        tk=yaksu(a);
        if(tk>0) printf("deficient number\n");
        else if(tk==0) printf("perfect number\n");
        else printf("abundant number\n");
        }
}
### Prompt

Create a solution in CPP for 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>
#include<cmath>
int yaksu(int a){
        if (a==1) return 1;
        int res=a;
        for(int i=sqrt(a);i>0;i--){
                if(a%i==0){
                        res-=i;
                        if(i!=a/i&&i!=1) res=res-a/i;
                }
                if(res<0) break;
        }
        return res;
}

int main(){
        int a,tk;
        while(1){
        scanf("%d",&a);
        if(a==0) return 0;
        tk=yaksu(a);
        if(tk>0) printf("deficient number\n");
        else if(tk==0) printf("perfect number\n");
        else printf("abundant number\n");
        }
}
```