#include <bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
#define repb(i,a,b) for(int i=a;i>=b;i--)
#define all(a) a.begin(),a.end()
#define o(a) cout<<a<<endl
#define int long long
#define fi first
#define se second
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;

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


signed main(){
    int n;
    while(cin >> n, n){
        int sum = calc(n);
        //o(sum);
        if(sum == n) cout << "perfect number" << endl;
        if(sum < n) cout << "deficient number" << endl;
        if(sum > n) cout << "abundant number" << endl;
    }
}
### Prompt

Develop a solution in Cpp to the problem described below:

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>
#define rep(i,a,n) for(int i=a;i<n;i++)
#define repb(i,a,b) for(int i=a;i>=b;i--)
#define all(a) a.begin(),a.end()
#define o(a) cout<<a<<endl
#define int long long
#define fi first
#define se second
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;

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


signed main(){
    int n;
    while(cin >> n, n){
        int sum = calc(n);
        //o(sum);
        if(sum == n) cout << "perfect number" << endl;
        if(sum < n) cout << "deficient number" << endl;
        if(sum > n) cout << "abundant number" << endl;
    }
}
```