#include <bits/stdc++.h>
using namespace std;

#define REP(i,n) for(int i=0;i<(n);i++)
#define REAP(i,a,n) for(int i=(a);i<(n);i++)
#define YES cout<<"Yes"<<endl
#define NO cout<<"No"<<endl
#define fr first
#define sc second
#define pb push_back
#define All(v) v.begin(),v.end()
typedef long long ll;

ll counum(ll n){
    ll cou=0;
    for(ll i=1;i*i<=n;i++){
        if(n%i==0){
            cou+=i;
            if(n/i!=i&&i!=1)cou+=n/i;
        }
    }
    if(n!=1)return cou;
    else return 0;
}

int main(){
    int n;
    while(cin >> n,n){
        ll cou=counum(n);
        if(n==cou)cout << "perfect number" << endl;
        else if(n<cou)cout << "abundant number" << endl;
        else cout << "deficient number" << endl;
    }

}


### Prompt

Please create a solution in CPP 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 <bits/stdc++.h>
using namespace std;

#define REP(i,n) for(int i=0;i<(n);i++)
#define REAP(i,a,n) for(int i=(a);i<(n);i++)
#define YES cout<<"Yes"<<endl
#define NO cout<<"No"<<endl
#define fr first
#define sc second
#define pb push_back
#define All(v) v.begin(),v.end()
typedef long long ll;

ll counum(ll n){
    ll cou=0;
    for(ll i=1;i*i<=n;i++){
        if(n%i==0){
            cou+=i;
            if(n/i!=i&&i!=1)cou+=n/i;
        }
    }
    if(n!=1)return cou;
    else return 0;
}

int main(){
    int n;
    while(cin >> n,n){
        ll cou=counum(n);
        if(n==cou)cout << "perfect number" << endl;
        else if(n<cou)cout << "abundant number" << endl;
        else cout << "deficient number" << endl;
    }

}


```