#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> mp;
#define inf 1e9

ll divisors(ll n){
    ll res = 0;
    for(ll i=1;i*i<=n;i++){
	if( n%i != 0) continue;
	res += i;
	if(n/i==i) continue;
	res += n/i;
    }
    return res-n;
}

int main(){
    while(1){
	ll n ;
	cin>>n;
	if(n==0) break;
	ll s = divisors(n);
	if( n==s){
	    cout<<"perfect number"<<endl;
	}else if( n > s){
	    cout<<"deficient number"<<endl;
	}else{
	    cout<<"abundant number"<<endl;
	}
    }
	return 0;
}


### Prompt

Construct a Cpp code solution to the problem outlined:

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>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> mp;
#define inf 1e9

ll divisors(ll n){
    ll res = 0;
    for(ll i=1;i*i<=n;i++){
	if( n%i != 0) continue;
	res += i;
	if(n/i==i) continue;
	res += n/i;
    }
    return res-n;
}

int main(){
    while(1){
	ll n ;
	cin>>n;
	if(n==0) break;
	ll s = divisors(n);
	if( n==s){
	    cout<<"perfect number"<<endl;
	}else if( n > s){
	    cout<<"deficient number"<<endl;
	}else{
	    cout<<"abundant number"<<endl;
	}
    }
	return 0;
}


```