#include <cstdio>
#include <utility>
#include <vector>
using namespace std;
typedef long long ll;
typedef pair<ll,int> P;
#define pb push_back
#define fi first
#define sec second
ll pow(ll a,int x)
{
	ll res = 1ll;
	while(x)
	{
		if(x&1)res*=a;
		a=a*a;
		x>>=1;
	}
	return res;
}
int main()
{
	while(1)
	{
		ll n,N;
		vector<P> fac;
		scanf("%lld",&n);
		if(n==0)break;
		N=n;
		for(ll i=2;i*i<=n;i++)
		{
			int cnt=0;
			while(n%i==0)
			{
				n/=i;
				cnt++;
			}
			if(cnt)fac.pb(P(i,cnt));
		}
		if(n!=1)fac.pb(P(n,1));
		ll S=1;
		for(int i=0;i<fac.size();i++)S *= (pow(fac[i].fi,fac[i].sec+1)-1)/(fac[i].fi-1);
		if(2*N==S)printf("perfect number\n");
		else if(2*N>S)printf("deficient number\n");
		else printf("abundant number\n");
	}
	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 <cstdio>
#include <utility>
#include <vector>
using namespace std;
typedef long long ll;
typedef pair<ll,int> P;
#define pb push_back
#define fi first
#define sec second
ll pow(ll a,int x)
{
	ll res = 1ll;
	while(x)
	{
		if(x&1)res*=a;
		a=a*a;
		x>>=1;
	}
	return res;
}
int main()
{
	while(1)
	{
		ll n,N;
		vector<P> fac;
		scanf("%lld",&n);
		if(n==0)break;
		N=n;
		for(ll i=2;i*i<=n;i++)
		{
			int cnt=0;
			while(n%i==0)
			{
				n/=i;
				cnt++;
			}
			if(cnt)fac.pb(P(i,cnt));
		}
		if(n!=1)fac.pb(P(n,1));
		ll S=1;
		for(int i=0;i<fac.size();i++)S *= (pow(fac[i].fi,fac[i].sec+1)-1)/(fac[i].fi-1);
		if(2*N==S)printf("perfect number\n");
		else if(2*N>S)printf("deficient number\n");
		else printf("abundant number\n");
	}
	return 0;
}
```