#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

#define MOD 1000000007
#define REP(i, N) for (int i = 0; i < N; ++i)
#define REP1(i, N) for (int i = 1; i <= N; ++i)
#define RREP(i, N) for (int i = N - 1; i >= 0; --i)
#define ALL(a) a.begin(), a.end()

int main() {
  ll n;
  while (cin >> n && n) {
    ll sum = 1;
    for (ll i = 2; i * i <= n; ++i) {
      if (n % i == 0) {
        sum += i;
        if (i != n / i) sum += n / i;
      }
    }
    if (n == 1) {
      cout << "deficient number" << endl;
    } else if (n == sum) {
      cout << "perfect number" << endl;
    } else if (n > sum) {
      cout << "deficient number" << endl;
    } else {
      cout << "abundant number" << endl;
    }
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 <algorithm>
#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

#define MOD 1000000007
#define REP(i, N) for (int i = 0; i < N; ++i)
#define REP1(i, N) for (int i = 1; i <= N; ++i)
#define RREP(i, N) for (int i = N - 1; i >= 0; --i)
#define ALL(a) a.begin(), a.end()

int main() {
  ll n;
  while (cin >> n && n) {
    ll sum = 1;
    for (ll i = 2; i * i <= n; ++i) {
      if (n % i == 0) {
        sum += i;
        if (i != n / i) sum += n / i;
      }
    }
    if (n == 1) {
      cout << "deficient number" << endl;
    } else if (n == sum) {
      cout << "perfect number" << endl;
    } else if (n > sum) {
      cout << "deficient number" << endl;
    } else {
      cout << "abundant number" << endl;
    }
  }
  return 0;
}

```