#include <cstdio>
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

int main() {

	while (1) {
		long long n;
		cin >> n;
		if (n == 0) break;
		int sum = 1;
		for (int i = 2; i * i <= n; i++) {
			if (n % i == 0) {
				sum += i;
				if(i != n/i)
					sum += n / i;
			}
		}
		if (n == 1) sum = 0;
		string s;
		if (sum == n) s = "perfect number";
		else if (sum > n) s = "abundant number";
		else s = "deficient number";

		cout << s << endl;
	}

	return 0;
}

### Prompt

Please create a solution in CPP 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 <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

int main() {

	while (1) {
		long long n;
		cin >> n;
		if (n == 0) break;
		int sum = 1;
		for (int i = 2; i * i <= n; i++) {
			if (n % i == 0) {
				sum += i;
				if(i != n/i)
					sum += n / i;
			}
		}
		if (n == 1) sum = 0;
		string s;
		if (sum == n) s = "perfect number";
		else if (sum > n) s = "abundant number";
		else s = "deficient number";

		cout << s << endl;
	}

	return 0;
}

```