#include <cstdio>
#include <vector>
#include <iostream>
using namespace std;

int main(int argc, char **argv)
{
    while(true) {
        int n;scanf("%d",&n);
        if(!n)return 0;
        int x=n;
        int prod=1;
        for(int i = 2; x>1; i++) {
            if(i*i>x)i=x;
            int num=1;
            int sum=1;
            while(x%i==0) {
                sum+=num*=i;
                x/=i;
            }
            prod *= sum;
        }
        n*=2;
        puts(prod==n?"perfect number":prod<n?"deficient number":"abundant number");
    }
    return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 <vector>
#include <iostream>
using namespace std;

int main(int argc, char **argv)
{
    while(true) {
        int n;scanf("%d",&n);
        if(!n)return 0;
        int x=n;
        int prod=1;
        for(int i = 2; x>1; i++) {
            if(i*i>x)i=x;
            int num=1;
            int sum=1;
            while(x%i==0) {
                sum+=num*=i;
                x/=i;
            }
            prod *= sum;
        }
        n*=2;
        puts(prod==n?"perfect number":prod<n?"deficient number":"abundant number");
    }
    return 0;
}
```