#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <utility>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;

#define rep(i,n) for(int i=0; i<n; i++)
typedef long long ll;
typedef pair<int,int> pii;

int main() {
	ll N;
	while(cin >> N, N) {
		ll cnt = (N==1 ? 0 : 1);
		for(int i=2; i*i<=N; i++) {
			if(N%i == 0) {
				cnt += i + (N/i);
				if(i == N/i) cnt -= i;
			}
		}
		if(cnt == N) {
			cout << "perfect number" << endl;
		}
		else if(cnt < N) {
			cout << "deficient number" << endl;
		}
		else {
			cout << "abundant number" << endl;
		}
	}
	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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <utility>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;

#define rep(i,n) for(int i=0; i<n; i++)
typedef long long ll;
typedef pair<int,int> pii;

int main() {
	ll N;
	while(cin >> N, N) {
		ll cnt = (N==1 ? 0 : 1);
		for(int i=2; i*i<=N; i++) {
			if(N%i == 0) {
				cnt += i + (N/i);
				if(i == N/i) cnt -= i;
			}
		}
		if(cnt == N) {
			cout << "perfect number" << endl;
		}
		else if(cnt < N) {
			cout << "deficient number" << endl;
		}
		else {
			cout << "abundant number" << endl;
		}
	}
	return 0;
}
```