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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define FOR(var, begin, end) for (int var = (begin); var <= (end); var++)
#define RFOR(var, begin, end) for (int var = (begin); var >= (end); var--)
#define REP(var, length) FOR(var, 0, length - 1)
#define RREP(var, length) RFOR(var, length - 1, 0)
#define EACH(value, var) for (auto value : var)
#define SORT(var) sort(var.begin(), var.end())
#define REVERSE(var) reverse(var.begin(), var.end())
#define RSORT(var) SORT(var); REVERSE(var)
#define OPTIMIZE_STDIO ios::sync_with_stdio(false);cin.tie(0);cout.precision(10);cout << fixed
#define endl '\n'

const ll MOD = 1e9 + 7;
struct mint {
    ll x;
    mint() : x(0) {}
    mint(ll x) : x((x % MOD + MOD) % MOD) {}
    mint& operator+=(const mint& r) { if ((x += r.x) >= MOD) { x -= MOD; } return *this; }
    mint& operator-=(const mint& r) { if ((x -= r.x) < 0) { x += MOD; } return *this; }
    mint& operator*=(const mint& r) { if ((x *= r.x) > MOD) { x %= MOD; } return *this; }
    mint& operator++() { if ((++x) >= MOD) { x -= MOD; } return *this; }
    mint operator++(int) { mint ret = x; if ((++x) >= MOD) { x-= MOD; } return ret; }
    mint& operator--() { if ((--x) < 0) { x += MOD; } return *this; }
    mint operator--(int) { mint ret = x; if ((--x) < 0) { x += MOD; } return ret; }
    mint operator+(const mint& r) { mint ret; return ret = x + r.x; }
    mint operator-(const mint& r) { mint ret; return ret = x - r.x; }
    mint operator*(const mint& r) { mint ret; return ret = x * r.x; }
    mint operator-() { return mint() - *this; }
    bool operator<(const mint& a) { return x < a.x; }
    bool operator>(const mint& a) { return x > a.x; }
    bool operator<=(const mint& a) { return x <= a.x; }
    bool operator>=(const mint& a) { return x >= a.x; }
    bool operator==(const mint& a) { return x == a.x; }
    bool operator!=(const mint& a) { return x != a.x; }
};
ostream& operator<<(ostream& os, const mint& r) { os << r.x; return os; }
ostream& operator>>(ostream& os, const mint& r) { os >> r.x; return os; }

void solve(istream& cin, ostream& cout) {
  int N;
  cin >> N;
  vector<int> A(N);
  REP(i, N) {
    cin >> A.at(i);
  }
  bool flag = true;
  REP(i, N) {
    if ((A.at(i) % 2 == 0)) {
      if (!(A.at(i) % 3 == 0 || A.at(i) % 5 == 0))
        flag = false;
    }
  }
  if (flag)
    cout << "APPROVED" << endl;
  else
    cout << "DENIED" << endl;
}

#ifndef TEST
int main() {
  OPTIMIZE_STDIO;
  solve(cin, cout);
}
#endif