#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
#include <numeric>
#include <functional>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
using namespace std;
typedef istringstream ISS;
typedef ostringstream OSS;
typedef vector<string> VS;
typedef int INT;
typedef vector<INT> VI;
typedef vector<VI> VVI;
typedef pair<INT, INT> II;
typedef long long LL;
const string OUT[3] = {
	"perfect number",
	"deficient number",
	"abundant number"
};
LL N;
bool input() {
	return cin >> N && N;
}
int solve() {
	LL sum = 0;
	for ( LL i = 1; i * i < N; ++ i ) {
		if ( N % i == 0 ) {
			if ( i != N / i && i != 1 ) {
				sum += i + N / i;
			} else {
				sum += i;
			}
		}
	}
	if ( N == sum ) return 0;
	return N < sum ? 2 : 1;
}
void output( int result ) {
	cout << OUT[result] << endl;
}
int main() {
    while ( input() ) {
        output( solve() );
    }
}
### 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 <sstream>
#include <fstream>
#include <iomanip>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
#include <numeric>
#include <functional>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
using namespace std;
typedef istringstream ISS;
typedef ostringstream OSS;
typedef vector<string> VS;
typedef int INT;
typedef vector<INT> VI;
typedef vector<VI> VVI;
typedef pair<INT, INT> II;
typedef long long LL;
const string OUT[3] = {
	"perfect number",
	"deficient number",
	"abundant number"
};
LL N;
bool input() {
	return cin >> N && N;
}
int solve() {
	LL sum = 0;
	for ( LL i = 1; i * i < N; ++ i ) {
		if ( N % i == 0 ) {
			if ( i != N / i && i != 1 ) {
				sum += i + N / i;
			} else {
				sum += i;
			}
		}
	}
	if ( N == sum ) return 0;
	return N < sum ? 2 : 1;
}
void output( int result ) {
	cout << OUT[result] << endl;
}
int main() {
    while ( input() ) {
        output( solve() );
    }
}
```