#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <cctype>
#include <numeric>
using namespace std;

#define rep(i,n) for(int (i)=0; (i)<(int)(n); ++(i))
#define foreach(c,i) for(__typeof((c).begin()) i=(c).begin();i!=(c).end();i++)

const string blah[] = { "deficient", "perfect", "abundant" };
int N;

int f() {
  vector<int> v;
  for (int i = 1; i * i <= N; ++i) {
    if (N % i == 0) {
      v.push_back(i);
      if (i != N/i)
        v.push_back(N/i);
    }
  }
  const int S = accumulate(v.begin(), v.end(), 0) - N;
  // printf("%d %d : ", N, S);
  return N == S ? 1 : N > S ? 0 : 2 ;
}

int main() {
  while (-1 != scanf("%d", &N)) {
    if (!N) break;
    printf("%s number\n", blah[f()].c_str());
  }
  return 0;
}
### 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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <cctype>
#include <numeric>
using namespace std;

#define rep(i,n) for(int (i)=0; (i)<(int)(n); ++(i))
#define foreach(c,i) for(__typeof((c).begin()) i=(c).begin();i!=(c).end();i++)

const string blah[] = { "deficient", "perfect", "abundant" };
int N;

int f() {
  vector<int> v;
  for (int i = 1; i * i <= N; ++i) {
    if (N % i == 0) {
      v.push_back(i);
      if (i != N/i)
        v.push_back(N/i);
    }
  }
  const int S = accumulate(v.begin(), v.end(), 0) - N;
  // printf("%d %d : ", N, S);
  return N == S ? 1 : N > S ? 0 : 2 ;
}

int main() {
  while (-1 != scanf("%d", &N)) {
    if (!N) break;
    printf("%s number\n", blah[f()].c_str());
  }
  return 0;
}
```