Problem:
Score: 200 points

Problem Statement

You are an immigration officer in the Kingdom of AtCoder. The document carried by an immigrant has some number of integers written on it, and you need to check whether they meet certain criteria.

According to the regulation, the immigrant should be allowed entry to the kingdom if and only if the following condition is satisfied:

All even numbers written on the document are divisible by 3 or 5.

If the immigrant should be allowed entry according to the regulation, output APPROVED; otherwise, print DENIED.

Notes

The condition in the statement can be rephrased as "If x is an even number written on the document, x is divisible by 3 or 5".
Here "if" and "or" are logical terms.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq A_i \leq 1000

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \dots A_N

Output

If the immigrant should be allowed entry according to the regulation, print APPROVED; otherwise, print DENIED.

Sample Input 1

5
6 7 9 10 31

Sample Output 1

APPROVED

The even numbers written on the document are 6 and 10.

All of them are divisible by 3 or 5, so the immigrant should be allowed entry.

Sample Input 2

3
28 27 24

Sample Output 2

DENIED

28 violates the condition, so the immigrant should not be allowed entry.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#pragma GCC optimize("Ofast")
//*********************************************DO IT NOW****************************************************************
#include<bits/stdc++.h> 
#include<random>
#define int long long   
#define pp pair<int,int> 
#define ss second 
#define ff first 
#define pb push_back
#define mod 1000000007
#define pi 3.14159265359
#define mk(arr,n,type)  type *arr=new type[n];
#define sl s1.length();
#define yes cout<< "YES"<<endl
#define no  cout<< "NO"<<endl
#define all(v) (v).begin(),(v).end()
#define s(v)   sort(v,v+n)
#define  mt mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define read(a,n) for(int i=0;i<n;i++)cin>>a[i]
#define print(a,n) for(int i=0;i<n;i++)cout<<a[i]<<" "
using namespace std;
//**********************************************DO IT NOW***************************************************************
void fastio()
{  
  #ifndef ONLINE_JUDGE
  freopen("INPUT.txt","r",stdin);
  freopen("OUTPUT.txt","w",stdout); 
  #endif
  ios_base :: sync_with_stdio(false); 
  cin.tie(NULL) ; 
  cout.tie(NULL);
}
int dp[101][1001],ans1=1e9,dp1[1000001];
int num[10];
int a[1000000];
int prime(int n)
{
 for(int i=2;i*i<=n;i++)
      {
          if(n%i==0)
          return 0;
      }
      return 1;
}
signed main()
{   
    fastio();
    int t=1;
   // cin>>t;
    while(t--)
    {  
        int n;
        cin>>n;
        int a[n];
        read(a,n);
        for(int i=0;i<n;i++)
        {
            if(a[i]%2==0)
            {
                if(a[i]%3==0 ||a[i]%5==0)
                {
                    cout<<"APPROVED";
                    return 0;
                }
                else
                {
                    cout<<"DENIED";
                    return 0;
                }
            }
        }
    }
    
}
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