#include<cstdio>
#include<set>
#include<cmath>
#include<iostream>
using namespace std;

int main()
{
    int n;
    while(~scanf("%d",&n),n)
    {
        set<int> s;
        s.insert(1);
        for(int i=2;i<sqrt(1.0*n)+2;i++)
        {
            if(n%i == 0)
            {
                s.insert(i);
                s.insert(n/i);
            }
        }
        set<int>::iterator p;
        int sum = 0;
        for(p=s.begin();p!=s.end();p++)
        {
            //cout<<*p<<endl;
            sum += *p;
        }
        if(n<=4) printf("deficient number\n");
        else{
             if(sum > n) printf("abundant number\n");
             else if(sum == n) printf("perfect number\n");
             else printf("deficient number\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<cstdio>
#include<set>
#include<cmath>
#include<iostream>
using namespace std;

int main()
{
    int n;
    while(~scanf("%d",&n),n)
    {
        set<int> s;
        s.insert(1);
        for(int i=2;i<sqrt(1.0*n)+2;i++)
        {
            if(n%i == 0)
            {
                s.insert(i);
                s.insert(n/i);
            }
        }
        set<int>::iterator p;
        int sum = 0;
        for(p=s.begin();p!=s.end();p++)
        {
            //cout<<*p<<endl;
            sum += *p;
        }
        if(n<=4) printf("deficient number\n");
        else{
             if(sum > n) printf("abundant number\n");
             else if(sum == n) printf("perfect number\n");
             else printf("deficient number\n");
        }
    }
    return 0;
}
```