// yukicoder: No.2 素因数ゲーム
// 2019.4.22 bal4u
// 素因数分解＋Nim

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

//// 素数テスト
long long calc_pow(int a, int k, int n)
{
	long long   p;
	unsigned bit;

	p = 1;
	for (bit = 0x80000000U; bit > 0; bit >>= 1) {
		if (p > 1) p = (p*p) % n;
		if (k & bit) p = (p*a) % n;
	}
	return p % n;
}

int suspect(int b, unsigned n)
{
	unsigned i;
	unsigned t, u, n1;
	long long x0, x1;

	u = n1 = n - 1;
	t = 0;
	while ((u & 1) == 0) {
		t++;
		u >>= 1;
	}
	x0 = calc_pow(b, u, n);
	for (i = 1; i <= t; i++) {
		x1 = (x0*x0) % n;
		if (x1 == 1 && x0 != 1 && x0 != n1) return 0;
		x0 = x1;
	}
	return x1 == 1;
}

int prime_test(int n)
{
	if (n == 1) return 0;          /* 1 */
	if (n == 2 || n == 3) return 1;          /* 2, 3 */
	if ((n & 1) == 0) return 0;  /* other even integers */
	if (!suspect(2, n)) return 0;
	if (n <= 1000000) {
		if (!suspect(3, n)) return 0;
	}
	else {
		if (!suspect(7, n)) return 0;
		if (!suspect(61, n)) return 0;
	}
	return 1;
}

int ptbl[] = {
		 3,   5,   7,  11,  13,  17,  19,  23,  29,
   31,  37,  41,  43,  47,  53,  59,  61,  67,  71,
   73,  79,  83,  89,  97, 101, 103, 107, 109, 113,
  127, 131, 137, 139, 149, 151, 157, 163, 167, 173,
  179, 181, 191, 193, 197, 199, 211, 223, 227, 229,
  233, 239, 241, 251, 257, 263, 269, 271, 277, 281,
  283, 293, 307, 311, 313, 317, 331, 337, 347, 349,
  353, 359, 367, 373, 379, 383, 389, 397, 401, 409,
  419, 421, 431, 433, 439, 443, 449, 457, 461, 463,
  467, 479, 487, 491, 499, 503, 509, 521, 523, 541,
  547, 557, 563, 569, 571, 577, 587, 593, 599, 601,
  607, 613, 617, 619, 631, 641, 643, 647, 653, 659,
  661, 673, 677, 683, 691, 701, 709, 719, 727, 733,
  739, 743, 751, 757, 761, 769, 773, 787, 797, 809,
  811, 821, 823, 827, 829, 839, 853, 857, 859, 863,
  877, 881, 883, 887, 907, 911, 919, 929, 937, 941,
  947, 953, 967, 971, 977, 983, 991, 997, 0 };

int xor_power(int n)
{
	int i, k, b, p, ans;

	k = 0, ans = 0, p = 0;
	if ((n & 1) == 0) {
		do n >>= 1, p++;
		while ((n & 1) == 0);
		ans ^= p, p = 0;
	}
	for (i = 0; n > 1 && (b=ptbl[i]); i++) {
		if (n % b) continue;
		do n /= b, p++;
		while (n % b == 0);
		ans ^= p, p = 0;
	}
	if (n > 1) {
		b = (int)sqrt((double)n);
		for (i = 1009; n > 1; i += 2) {
			if (i > b) { ans ^= 1; break; }
			if (n % i == 0) {
				do n /= i, p++;
				while (n % i == 0);
				ans ^= p;
			}
		}
	}
	return ans;
}

int main()
{
	int N, ans;
	scanf("%d", &N);
	if (prime_test(N)) ans = 1;
	else ans = xor_power(N);
	puts(ans ? "Alice" : "Bob");
	return 0;
}