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;

using ll = long long;
#define repi(n) for (int i = 0; i < (n); i++)
#define repj(n) for (int j = 0; j < (n); j++)
#define repk(n) for (int k = 0; k < (n); k++)
#define repl(n) for (int l = 0; l < (n); l++)
#define rep(i, n) for (int i = 0; (i) < (n); i++)
#define repr(i, a, b) for (auto i = (a); i < (b); i++)
#define repv(itr) for (auto&& v : (itr))
#define repv2(itr) for (auto&& v2 : (itr))
#define repv3(itr) for (auto&& v3 : (itr))
#define updatemax(t, v) (t = max((t), (v)))
#define updatemin(t, v) (t = min((t), (v)))
const int dx[] = {-1, 0, 0, 1, -1, -1, 1, 1};
const int dy[] = {0, -1, 1, 0, -1, 1, -1, 1};
const double PI = atan(1.0) * 4;
template <typename T>
T minptr(T begin, T end) {
  T re = begin;
  for (T i = begin + 1; i != end; i++) {
    if (*i < *re) re = i;
  }
  return re;
}
template <typename T>
T maxptr(T begin, T end) {
  T re = begin;
  for (T i = begin + 1; i != end; i++) {
    if (*i > *re) re = i;
  }
  return re;
}
int __vmax(int x) {
  return INT_MAX;
}
double __vmax(double x) {
  return 1e+300;
}
ll __vmax(ll x) {
  return LLONG_MAX;
}
int __vmin(int x) {
  return INT_MIN;
}
double __vmin(double x) {
  return -1e+300;
}
ll __vmin(ll x) {
  return LLONG_MIN;
}
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : b == 1 ? 1 : gcd(b, a % b);
}
template <typename T>
T bsearch(T begin, T end, function<bool(T)> f, T targetwidth) {
  if (f(begin)) return begin;
  if (!f(end)) return end;
  while (abs(end - begin) > targetwidth) {
    T m = (begin + end) / 2;
    if (f(m)) {
      end = m;
    } else {
      begin = m;
    }
  }
  return end;
}
ll bsearchi(ll begin, ll end, function<bool(ll)> f) {
  return bsearch<ll>(begin, end, f, 1);
}
double bsearchd(double begin, double end, function<bool(double)> f) {
  return bsearch<double>(begin, end, f, max(abs(begin), abs(end)) * 1e-14);
}
ll modpow(ll a, ll b, ll m) {
  ll re = 1, k = 1;
  while (k <= b) {
    if (b & k) {
      re *= a;
      re %= m;
    }
    k = k << 1;
    a *= a;
    a %= m;
  }
  return re;
}
ll modinv(ll a, ll m) {
  ll b = m, u = 1, v = 0;
  while (b) {
    ll t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0) u += m;
  return u;
}
ll modbinomial(ll n, ll k, ll m) {
  k = min(k, n - k);
  if (k < 0) return 0;
  ll re = 1;
  for (ll i = 0; i < k; i++) {
    re *= n - i;
    re %= m;
    re *= modinv(i + 1, m);
    re %= m;
  }
  return re;
}
template <typename T>
vector<T> lis(T begin, T end, bool allowequal = false, bool lds = false) {
  using V = typename iterator_traits<T>::value_type;
  int n = end - begin;
  vector<V> a(n, lds ? __vmin(*begin) : __vmax(*begin));
  vector<int> id(n);
  if (lds && allowequal) {
    for (int i = 0; i < n; i++) {
      id[i] = n - 1 -
              (lower_bound(a.rbegin(), a.rend(), begin[i]) - 1 - a.rbegin());
      a[id[i]] = begin[i];
    }
  } else if (lds) {
    for (int i = 0; i < n; i++) {
      id[i] = n - 1 -
              (upper_bound(a.rbegin(), a.rend(), begin[i]) - 1 - a.rbegin());
      a[id[i]] = begin[i];
    }
  } else if (allowequal) {
    for (int i = 0; i < n; i++) {
      id[i] = upper_bound(a.begin(), a.end(), begin[i]) - a.begin();
      a[id[i]] = begin[i];
    }
  } else {
    for (int i = 0; i < n; i++) {
      id[i] = lower_bound(a.begin(), a.end(), begin[i]) - a.begin();
      a[id[i]] = begin[i];
    }
  }
  int m = *maxptr(id.begin(), id.end());
  vector<T> re(m + 1);
  for (int i = n - 1; i >= 0; i--) {
    if (id[i] == m) re[m--] = begin + i;
  }
  return re;
}
class uftree {
 private:
  int n, groups;
  vector<int> parent;

 public:
  uftree(int _n) {
    groups = n = _n;
    parent.resize(n);
    for (int i = 0; i < n; i++) parent[i] = i;
  }
  int group(int x) {
    if (parent[x] == x) return x;
    return parent[x] = group(parent[x]);
  }
  void merge(int x, int y) {
    int gx = group(x), gy = group(y);
    if (gx != gy) {
      parent[gx] = gy;
      groups--;
    }
  }
  int countGroups() { return groups; }
};
template <typename T>
class segtree {
 private:
  int n;
  function<T(T, T)> f;
  T e;
  vector<T> data;
  void _updateP(int i) {
    data[i] = f(data[i * 2 + 1], data[i * 2 + 2]);
    if (i) _updateP((i - 1) / 2);
  }
  T _calc(int begin, int end, int node, int nodeBegin, int nodeEnd) {
    if (end <= nodeBegin || nodeEnd <= begin) {
      return e;
    } else if (begin <= nodeBegin && nodeEnd <= end) {
      return data[node];
    } else {
      int m = (nodeBegin + nodeEnd) / 2;
      T left = _calc(begin, end, node * 2 + 1, nodeBegin, m);
      T right = _calc(begin, end, node * 2 + 2, m, nodeEnd);
      return f(left, right);
    }
  }

 public:
  segtree(int n_, function<T(T, T)> f_, T e_, T fill) {
    n = pow(2, ceil(log2(n_)));
    f = f_;
    e = e_;
    data.resize(n * 2 - 1);
    for (int i = 0; i < n * 2 - 1; i++) data[i] = fill;
  }
  segtree(int n_, function<T(T, T)> f_, T e_) : segtree(n_, f_, e_, e_) {}
  T value(int i) { return data[n - 1 + i]; }
  void update(int i, T value) {
    data[n - 1 + i] = value;
    if (n) _updateP((n - 2 + i) / 2);
  }
  T calc(int begin, int end) { return _calc(begin, end, 0, 0, n); }
  static T max(T a, T b) { return std::max(a, b); }
  static T min(T a, T b) { return std::min(a, b); }
  static T sum(T a, T b) { return a + b; }
};

int a[100];
main() {
  int n;
  cin >> n;
  repi(n) cin >> a[i];
  bool p = true;
  repi(n) {
    if (a[i] % 2 == 0 && a[i] % 3 != 0 && a[i] % 5 != 0) p = false;
  }
  cout << (p ? "APPROVED" : "DENIED") << endl;
}