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

typedef long long ll;

//不足:-1 完全:0 過剰:1
int judge(ll n){
  ll sum=0;
  for(ll i=1; i*i<=n; i++){
    if(n%i==0)sum+=(i+n/i);
    if(i==1)sum-=n;
    if(i*i==n)sum-=i;
  }
  if(sum>n)return 1;
  if(sum<n)return -1;
  return 0;
}

int main(){
  ll n;
  while(true){
    cin >> n;
    if(n==0)break;
    switch(judge(n)){
    case -1 : cout << "deficient number" << endl; break;
    case 0 : cout << "perfect number" << endl; break;
    case 1 : cout << "abundant number" << endl; break;
    }
  }

  return 0;
}
### 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
//Perfect Number
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

//不足:-1 完全:0 過剰:1
int judge(ll n){
  ll sum=0;
  for(ll i=1; i*i<=n; i++){
    if(n%i==0)sum+=(i+n/i);
    if(i==1)sum-=n;
    if(i*i==n)sum-=i;
  }
  if(sum>n)return 1;
  if(sum<n)return -1;
  return 0;
}

int main(){
  ll n;
  while(true){
    cin >> n;
    if(n==0)break;
    switch(judge(n)){
    case -1 : cout << "deficient number" << endl; break;
    case 0 : cout << "perfect number" << endl; break;
    case 1 : cout << "abundant number" << endl; break;
    }
  }

  return 0;
}
```