#include "bits/stdc++.h"
#include<unordered_map>
#include<unordered_set>
#pragma warning(disable:4996)
using namespace std;
using ld = long double;
const ld eps = 1e-9;

vector<long long int> divisor(long long int n) {
	vector<long long int> res;
	for (long long int i = 1; i * i <= n; ++i) {
		if (n % i == 0) {
			res.push_back(i);
			if (i * i != n) res.push_back(n / i);
		}
	}
	sort(begin(res), end(res));
	return res;
}

int main() {
	while (1) {
		long long int N; cin >> N;
		if (!N)break;
		auto v(divisor(N));
		long long int sum = accumulate(v.begin(), v.end(), 0ll);
		if (sum > 2 * N)cout << "abundant number" << endl;
		else if (sum == 2 * N)cout << "perfect number" << endl;
		else cout << "deficient number" << endl;

	}
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 "bits/stdc++.h"
#include<unordered_map>
#include<unordered_set>
#pragma warning(disable:4996)
using namespace std;
using ld = long double;
const ld eps = 1e-9;

vector<long long int> divisor(long long int n) {
	vector<long long int> res;
	for (long long int i = 1; i * i <= n; ++i) {
		if (n % i == 0) {
			res.push_back(i);
			if (i * i != n) res.push_back(n / i);
		}
	}
	sort(begin(res), end(res));
	return res;
}

int main() {
	while (1) {
		long long int N; cin >> N;
		if (!N)break;
		auto v(divisor(N));
		long long int sum = accumulate(v.begin(), v.end(), 0ll);
		if (sum > 2 * N)cout << "abundant number" << endl;
		else if (sum == 2 * N)cout << "perfect number" << endl;
		else cout << "deficient number" << endl;

	}
	return 0;
}
```