#include<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<n;i++)
using namespace std;

int main() {
	int n;
	while (scanf("%d", &n), n) {
		int cnt = 0;
		for (int i = 2; i*i <= n; i++) {
			if (n%i == 0) {
				cnt += i;
				if (n / i != i)cnt += n / i;
			}
		}
		if (n != 1)cnt += 1;
		if (cnt > n)puts("abundant number");
		else if (cnt == n)puts("perfect number");
		else puts("deficient number");
	}
}
### Prompt

Construct a CPP code solution to the problem outlined:

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,n)for(int i=0;i<n;i++)
using namespace std;

int main() {
	int n;
	while (scanf("%d", &n), n) {
		int cnt = 0;
		for (int i = 2; i*i <= n; i++) {
			if (n%i == 0) {
				cnt += i;
				if (n / i != i)cnt += n / i;
			}
		}
		if (n != 1)cnt += 1;
		if (cnt > n)puts("abundant number");
		else if (cnt == n)puts("perfect number");
		else puts("deficient number");
	}
}
```