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 <iostream>
#include <vector>
#include <string>
#include <numeric>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <queue>

using namespace std;
using ll = long long;
//const int MOD = 1000000007;

#define rep(countName, left, right, up) for (ll countName = left; countName < right; countName += up)
#define rev(countName, right, left, down) for (ll countName = right - 1; countName >= left; countName -= down)

/* ****************************** ユークリッド互除法 ****************************** */
unsigned long long GCD(unsigned long long a, unsigned long long b)
{
    if (a % b == 0)
    {
        return b;
    }
    else
    {
        return GCD(b, a % b);
    }
}

unsigned long long GCD(vector<unsigned long long> v)
{
    for (auto i = 0; i < (long long)v.size() - 1; i++)
    {
        v[i - 1] = GCD(v[i - 1], v[i]);
    }

    return v[(long long)v.size() - 1];
}
/* ****************************** ユークリッド互除法 ****************************** */

/* ****************************** 最小公倍数 ****************************** */

unsigned long long LCM(unsigned long long x, unsigned long long y)
{
    unsigned long long gcd = GCD(x, y);
    return (x / gcd * y);
}

unsigned long long LCM(vector<unsigned long long> v)
{
    for (long long i = 0; i < (long long)v.size() - 1; i++)
    {
        v[i - 1] = LCM(v[i - 1], v[i]);
    }

    return v[(long long)v.size() - 1];
}

/* ****************************** 最小公倍数 ****************************** */

/* ****************************** 累積和 ****************************** */
vector<long long> CUSUM(vector<long long> v)
{
    int n = v.size();
    vector<long long> cusum(n);
    cusum[0] = v[0];

    rep(i, 1, n, 1)
    {
        cusum[i] = v[i] + cusum[i - 1];
    }
    return cusum;
}

//部分和

long long partSum(vector<long long> &&v, long long left, long long right)
{
    return (v[right] - v[left]);
}

/* ****************************** 累積和 ****************************** */

/* ****************************** 二項係数 ****************************** */
//けんちょんさん作

const int MAX = 800000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit()
{
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k)
{
    if (n < k)
    {
        return 0;
    }
    if (n < 0 || k < 0)
    {
        return 0;
    }

    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

/* ****************************** 二項係数 ****************************** */

/* ****************************** 素数判定 ****************************** */

bool isPrime(long long n)
{
    if (n <= 1)
    {
        return false;
    }
    else if (n == 2)
    {
        return true;
    }
    else if (n % 2 == 0)
    {
        return false;
    }
    else
    {
        rep(i, 3, sqrt(n) + 1, 2)
        {
            if (n % i == 0)
            {
                return false;
            }
        }
        return true;
    }
}

/* ****************************** 二項係数 ****************************** */

/* ****************************** main関数 ****************************** */

string solve1()
{
    int n;
    string ans = "APPROVED";
    cin >> n;
    vector<int> a(n);
    
    rep(i, 0, n, 1)
    {
        cin >> a[i];

        if (a[i] % 2 == 0 && ans == "APPROVED")
        {
            if (!(a[i] % 3 == 0 || a[i] % 5 == 0))
            {
                ans = "DENIED";
            }
        }
    }

    return ans;
}

int main(void)
{
    cout << solve1() << endl;
    return 0;
}