#include<iostream>
#include<vector>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;
int main() {
	long long  a = 6;
	while (cin >> a&&a != 0) {
		long long sum = 1,a2=a;
		int i = 2;
		while (a != 1) {
			if (a%i == 0) {
				int j = 1;
				long long sum2 = 1;
				while (a%i == 0) {
					sum2 += (pow(i, j));
					j++;
					a /= i;
				}
				sum *= sum2;
			}
			if(i==2)
				i++;
			else i += 2;
		}
		sum -= a2;
		if (a2 == sum) cout << "perfect number" << endl;
		else if (a2 > sum) cout << "deficient number" << endl;
		else cout << "abundant number" << endl;
	}
}
### 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
#include<iostream>
#include<vector>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;
int main() {
	long long  a = 6;
	while (cin >> a&&a != 0) {
		long long sum = 1,a2=a;
		int i = 2;
		while (a != 1) {
			if (a%i == 0) {
				int j = 1;
				long long sum2 = 1;
				while (a%i == 0) {
					sum2 += (pow(i, j));
					j++;
					a /= i;
				}
				sum *= sum2;
			}
			if(i==2)
				i++;
			else i += 2;
		}
		sum -= a2;
		if (a2 == sum) cout << "perfect number" << endl;
		else if (a2 > sum) cout << "deficient number" << endl;
		else cout << "abundant number" << endl;
	}
}
```