#define _USE_MATH_DEFINES
#define INF 0x3f3f3f3f
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <limits>
#include <map>
#include <string>
#include <cstring>
#include <set>
#include <deque>
#include <bitset>
#include <list>
#include <cctype>
#include <utility>

using namespace std;

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

int tx[] = {0,1,0,-1};
int ty[] = {-1,0,1,0};

static const double EPS = 1e-8;

int main(){
	int n;

	while(~scanf("%d",&n)){
		if(n==0) break;

		int S = 0;
		for(int i=1;i*i<=n;i++){
			if(n % i == 0 && i != n){
				S+=i;
				if(n/i != i && n/i != n) S+=n/i;
			}
		}
		
		if(S < n){
			printf("deficient number\n");
		}
		else if(S==n){
			printf("perfect number\n");
		}
		else{
			printf("abundant number\n");
		}
	}
}
### Prompt

Please provide a cpp coded solution 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
#define _USE_MATH_DEFINES
#define INF 0x3f3f3f3f
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <limits>
#include <map>
#include <string>
#include <cstring>
#include <set>
#include <deque>
#include <bitset>
#include <list>
#include <cctype>
#include <utility>

using namespace std;

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

int tx[] = {0,1,0,-1};
int ty[] = {-1,0,1,0};

static const double EPS = 1e-8;

int main(){
	int n;

	while(~scanf("%d",&n)){
		if(n==0) break;

		int S = 0;
		for(int i=1;i*i<=n;i++){
			if(n % i == 0 && i != n){
				S+=i;
				if(n/i != i && n/i != n) S+=n/i;
			}
		}
		
		if(S < n){
			printf("deficient number\n");
		}
		else if(S==n){
			printf("perfect number\n");
		}
		else{
			printf("abundant number\n");
		}
	}
}
```