#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;
long long power(long x, unsigned long long y, long p)
{
    int res = 1;      // Initialize result
    x = x % p;  // Update x if it is more than or
                // equal to p
    while (y > 0)
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res*x) % p;
 
        // y must be even now
        y = y>>1; // y = y/2
        x = (x*x) % p;
    }
    return res;
}
 
// This function is called for all k trials. It returns
// false if n is composite and returns false if n is
// probably prime.
// d is an odd number such that  d*2<sup>r</sup> = n-1
// for some r >= 1
bool miillerTest(long long  d, long long n)
{
    // Pick a random number in [2..n-2]
    // Corner cases make sure that n > 4
    long long a = 2 + rand() % (n - 4);
 
    // Compute a^d % n
    long long x = power(a, d, n);
 
    if (x == 1  || x == n-1)
       return true;
 
    // Keep squaring x while one of the following doesn't
    // happen
    // (i)   d does not reach n-1
    // (ii)  (x^2) % n is not 1
    // (iii) (x^2) % n is not n-1
    while (d != n-1)
    {
        x = (x * x) % n;
        d *= 2; 
        if (x == 1)      return false;
        if (x == n-1)    return true;
    }
    // Return composite
    return false;
}
 
// It returns false if n is composite and returns true if n
// is probably prime.  k is an input parameter that determines
// accuracy level. Higher value of k indicates more accuracy.
bool isPrime(long long n, long long k)
{
    // Corner cases
    if (n <= 1 || n == 4)  return false;
    if (n <= 3) return true;
 
    // Find r such that n = 2^d * r + 1 for some r >= 1
    long long d = n - 1;
    while (d % 2 == 0)
        d /= 2;
 
    // Iterate given nber of 'k' times
    for (long  long i = 0; i < k; i++)
         if (miillerTest(d, n) == false)
              return false;
 
    return true;
}

int main(){
    int g;
    cin >> g;
    for(int a0 = 0; a0 < g; a0++){
        int n;
        cin >> n;
        // your code goes here
        
         int broj_p = 0;
        if(n >= 2) broj_p++;
        for(int i = 1; i <= n; i+=2)
            if(isPrime(i,25)) broj_p++;
       //     cout << broj_p<< "\n";
        if(broj_p%2 == 0) cout << "Bob\n";
        else cout << "Alice\n";
    }
    return 0;
}