#include <bits/stdc++.h>

#define ALL(a) (a).begin(), (a).end()
#define llong long long

using namespace std;

vector<llong> divisors(llong n){
	vector<llong> res;
	for(llong i = 1; i * i <= n; ++i){
		if(n % i != 0) continue;
		res.push_back(i);
		if(n/i == i) continue;
			
		res.push_back(n/i);
		
	}
	return res;
}

signed main(){

	while(true){
		llong n;
		cin >> n;
		
		if(n == 0) break;
		
		vector<llong> v;
		v = divisors(n);
		llong ans = 0;
		for(auto itr = v.begin(); itr!=v.end(); ++itr){
			//
			//
			//cout << *itr << " ";
			if(*itr != n)
				ans += *itr;
		}
		//cout << endl;
		if(ans == n){
			cout << "perfect number" << endl;
		}else if(ans < n){
			cout << "deficient number" << endl;
		}else{
			cout << "abundant number" << endl;
		}
	}
	return 0;
}



### Prompt

Develop a solution in CPP 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>

#define ALL(a) (a).begin(), (a).end()
#define llong long long

using namespace std;

vector<llong> divisors(llong n){
	vector<llong> res;
	for(llong i = 1; i * i <= n; ++i){
		if(n % i != 0) continue;
		res.push_back(i);
		if(n/i == i) continue;
			
		res.push_back(n/i);
		
	}
	return res;
}

signed main(){

	while(true){
		llong n;
		cin >> n;
		
		if(n == 0) break;
		
		vector<llong> v;
		v = divisors(n);
		llong ans = 0;
		for(auto itr = v.begin(); itr!=v.end(); ++itr){
			//
			//
			//cout << *itr << " ";
			if(*itr != n)
				ans += *itr;
		}
		//cout << endl;
		if(ans == n){
			cout << "perfect number" << endl;
		}else if(ans < n){
			cout << "deficient number" << endl;
		}else{
			cout << "abundant number" << endl;
		}
	}
	return 0;
}



```