Alice: "Hi, Bob! Let's play Nim!"
Bob: "Are you serious? I don't want to play it. I know how to win the game."
Alice: "Right, there is an algorithm to calculate the optimal move using XOR. How about changing the rule so that a player loses a game if he or she makes the XOR to $0$?"
Bob: "It sounds much better now, but I suspect you know the surefire way to win."
Alice: "Do you wanna test me?"
This game is defined as follows.


1. The game starts with $N$ heaps where the $i$-th of them consists of $a_i$ stones.
2. A player takes any positive number of stones from any single one of the heaps in one move.
3. Alice moves first. The two players alternately move.
4. If the XOR sum, $a_1$ XOR $a_2$ XOR $...$ XOR $a_N$, of the numbers of remaining stones of these heaps becomes $0$ as a result of a player's move, the player loses.



Your task is to find which player will win if they do the best move.



Input

The input consists of a single test case in the format below.


$N$
$a_1$
$\vdots$
$a_N$


The first line contains an integer $N$ which is the number of the heaps ($1 \leq N \leq 10^5$). Each of the following $N$ lines gives the number of stones in each heap ($1 \leq a_i \leq 10^9$).

Output

Output the winner, Alice or Bob, when they do the best move.

Examples

Input

2
1
1


Output

Alice


Input

5
1
2
3
4
5


Output

Bob


Input

10
1
2
3
4
5
6
7
8
9
10


Output

Alice
#include<bits/stdc++.h>
#define all(v) (v).begin(),(v).end()
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define ri(x) scanf("%d",&(x))
#define ri2(x,y) scanf("%d %d",&(x),&(y))
#define ri3(x,y,z) scanf("%d %d %d",&(x),&(y),&(z))
#define rll(x) scanf("%lld",&(x))
#define rll2(x,y) scanf("%lld %lld",&(x),&(y))
#define rll3(x,y,z) scanf("%lld %lld %lld",&(x),&(y),&(z))
#define gc(x) ((x) = getchar())
using namespace::std;

const long double PI = acos(-1);
const long long MOD = 1000000000 +7;

typedef long long ll;
typedef pair<ll,ll> pll;
typedef pair<ll,pll> tll;
typedef pair<int,int> ii;
typedef pair<int,ii> iii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<tll> vtll;
typedef vector<string> vs;
typedef set<int> si;
typedef set<ii> sii;
typedef set<iii> siii;

ll gcd(ll a, ll b){ return b==0?a:gcd(b,a%b);}

ll add(ll a, ll b, ll m = MOD){
	if(a >= m) a %= m;
	if(b >= m) b %= m;
	if(a < 0) a += m;
	if(b < 0) b += m;
	ll res = a+b;
	if(res >= m or res <= -m) res %= m;
	if(res < 0) res += m;
	return res;
}

ll mul(ll a, ll b, ll m = MOD){
	if(a >= m) a %= m;
	if(b >= m) b %= m;
	if(a < 0) a += m;
	if(b < 0) b += m;
	ll res = a*b;
	if(res >= m or res <= -m) res %= m;
	if(res < 0) res += m;
	return res;
}

ll pow_mod(ll a, ll b, ll m = MOD){
	ll res = 1LL;
	a = a%m;
	while(b){
		if(b&1) res = mul(res,a,m);
		b >>= 1;
		a = mul(a,a,m);
	}
	return res;
}

ll fastexp(ll a, ll b){
	ll res = 1LL;
	while(b){
		if(b&1) res = res*a;
		b >>= 1;
		a *= a;
	}
	return res;
}

int gcdExtendido(int a, int b, int *x, int *y){
	if(a == 0){
		*x = 0;
		*y = 1;
		return b;
	}
	int x1, y1;
	int gcd = gcdExtendido(b%a,a,&x1,&y1);
	
	*x = y1-(b/a)*x1;
	*y = x1;
	return gcd;
}

int modInverso(int a, int m){
	int x, y;
	int g = gcdExtendido(a,m,&x,&y);
	if(g!=1) return -1;
	else return (x%m + m)%m;
}

/****************************************
*************P*L*A*N*T*I*L*L*A************
*****************************************/

int n;

int main(){
	ri(n);
	int Xor = 0;
	int odds = 0;
	for(int i=0; i<n; i++){
		int x;
		ri(x);
		if(x & 1) odds += 1;
		Xor ^= x;
	}
	if(Xor == 1) puts("Bob");
	else if(Xor == 0){
		if(odds > 0) puts("Alice");
		else puts("Bob");
	}
	else puts("Alice");
	return 0;
}
