#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>

using namespace std;

#define ll long long
#define ld long double
#define EPS 0.0000000001
#define INF 1e9
#define MOD 1000000007
#define rep(i,n) for(i=0;i<n;i++)
#define loop(i,a,n) for(i=a;i<n;i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)

typedef vector<int> vi;
typedef pair<int,int> pii;

int main(void) {
  int i,j;
  int n;

  while(1){
    cin>>n;
    if(n==0)break;
    if(n==1){
      cout<<"deficient number"<<endl;
      continue;
    }
    int sum=1;
    for(i=2;i*i<=n;i++)
      if(n%i==0){
        sum+=i;
        if(i*i<n)sum+=n/i;
      }
    if(sum<n) cout<<"deficient number"<<endl;
    else if(sum==n) cout<<"perfect number"<<endl;
    else cout<<"abundant number"<<endl;
  }
}
### Prompt

Your task is to create 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<algorithm>
#include<cstdio>
#include<cmath>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>

using namespace std;

#define ll long long
#define ld long double
#define EPS 0.0000000001
#define INF 1e9
#define MOD 1000000007
#define rep(i,n) for(i=0;i<n;i++)
#define loop(i,a,n) for(i=a;i<n;i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)

typedef vector<int> vi;
typedef pair<int,int> pii;

int main(void) {
  int i,j;
  int n;

  while(1){
    cin>>n;
    if(n==0)break;
    if(n==1){
      cout<<"deficient number"<<endl;
      continue;
    }
    int sum=1;
    for(i=2;i*i<=n;i++)
      if(n%i==0){
        sum+=i;
        if(i*i<n)sum+=n/i;
      }
    if(sum<n) cout<<"deficient number"<<endl;
    else if(sum==n) cout<<"perfect number"<<endl;
    else cout<<"abundant number"<<endl;
  }
}
```