#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>

#define ll long long
using namespace std;
 
/* 
 * calculates (a * b) % c taking into account that a * b might overflow 
 */
ll mulmod(ll a, ll b, ll mod)
{
    ll x = 0,y = a % mod;
    while (b > 0)
    {
        if (b % 2 == 1)
        {    
            x = (x + y) % mod;
        }
        y = (y * 2) % mod;
        b /= 2;
    }
    return x % mod;
}
/* 
 * modular exponentiation
 */
ll modulo(ll base, ll exponent, ll mod)
{
    ll x = 1;
    ll y = base;
    while (exponent > 0)
    {
        if (exponent % 2 == 1)
            x = (x * y) % mod;
        y = (y * y) % mod;
        exponent = exponent / 2;
    }
    return x % mod;
}
 
/*
 * Miller-Rabin primality test, iteration signifies the accuracy
 */
bool Miller(ll p,int iteration)
{
    if (p < 2)
    {
        return false;
    }
    if (p != 2 && p % 2==0)
    {
        return false;
    }
    ll s = p - 1;
    while (s % 2 == 0)
    {
        s /= 2;
    }
    for (int i = 0; i < iteration; i++)
    {
        ll a = rand() % (p - 1) + 1, temp = s;
        ll mod = modulo(a, temp, p);
        while (temp != p - 1 && mod != 1 && mod != p - 1)
        {
            mod = mulmod(mod, mod, p);
            temp *= 2;
        }
        if (mod != p - 1 && temp % 2 == 0)
        {
            return false;
        }
    }
    return true;
}

int main(){
    int g;
    cin >> g;
    long long broj[100001];
    broj[1] = 0;
    for(int i = 2; i <= 100000; i++){
        if(Miller(i,25)){
            broj[i] = broj[i-1]+1;
        }else{
            broj[i] = broj[i-1];
        }
    }
    
    for(int a0 = 0; a0 < g; a0++){
        long long n;
        cin >> n;
        
        if(broj[n] % 2){
            cout << "Alice" << endl;
        }else{
            cout << "Bob" << endl;
        }
        // your code goes here
    }
    return 0;
}