#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define all(a)  (a).begin(),(a).end()
#define pb push_back
#define INF 999999999

vector<ll> divisor(ll n){
    vector<ll> ret;
    for(int i=1;i<=sqrt(n);i++){
        if(n%i==0){
            ret.pb(i);
            if(i!=n/i)ret.pb(n/i);
        }
    }
//  sort(all(ret));
    return ret;
}



int main(){
    int n;
    while(cin>>n&&n){
        vector<ll> res = divisor(n);
        int sum = 0;
        rep(i,res.size()){
            sum+=res[i];
        }
        sum-=n;
        
        if(sum==n)cout<<"perfect number"<<endl;
        if(n>sum)cout<<"deficient number"<<endl;
        if(n<sum)cout<<"abundant number"<<endl;
    }
}
### Prompt

Your challenge is to write 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 "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define all(a)  (a).begin(),(a).end()
#define pb push_back
#define INF 999999999

vector<ll> divisor(ll n){
    vector<ll> ret;
    for(int i=1;i<=sqrt(n);i++){
        if(n%i==0){
            ret.pb(i);
            if(i!=n/i)ret.pb(n/i);
        }
    }
//  sort(all(ret));
    return ret;
}



int main(){
    int n;
    while(cin>>n&&n){
        vector<ll> res = divisor(n);
        int sum = 0;
        rep(i,res.size()){
            sum+=res[i];
        }
        sum-=n;
        
        if(sum==n)cout<<"perfect number"<<endl;
        if(n>sum)cout<<"deficient number"<<endl;
        if(n<sum)cout<<"abundant number"<<endl;
    }
}
```