//テストケース100個以下、1 <= n <= 10^8
#include<iostream>
#include<cmath>
using namespace std;

char *out[3] = {"perfect number", "deficient number", "abundant number"};

int main() {
	int n;
	while( cin >> n ) {
		if (!n) break;
		
		int i, s = 0;
		for( i = 1; i <= (int)(sqrt((double)n) + 1e-6); i++ ) {
			if ( n % i == 0 ) {
				s += i;
				if ( n/i > i )
					s += n/i;
			}
		}
		s -= n;
		//cout << s << endl;
		
		if ( n == s )
			cout << out[0] << endl;
		if ( n > s )
			cout << out[1] << endl;
		if ( n < s )
			cout << out[2] << endl;
	}
	return 0;
}
### Prompt

In CPP, your task is to solve 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
//テストケース100個以下、1 <= n <= 10^8
#include<iostream>
#include<cmath>
using namespace std;

char *out[3] = {"perfect number", "deficient number", "abundant number"};

int main() {
	int n;
	while( cin >> n ) {
		if (!n) break;
		
		int i, s = 0;
		for( i = 1; i <= (int)(sqrt((double)n) + 1e-6); i++ ) {
			if ( n % i == 0 ) {
				s += i;
				if ( n/i > i )
					s += n/i;
			}
		}
		s -= n;
		//cout << s << endl;
		
		if ( n == s )
			cout << out[0] << endl;
		if ( n > s )
			cout << out[1] << endl;
		if ( n < s )
			cout << out[2] << endl;
	}
	return 0;
}
```