Alice and Bob are playing a game with nn piles of stones. It is guaranteed that nn is an even number. The ii-th pile has aiai stones.

Alice and Bob will play a game alternating turns with Alice going first.

On a player's turn, they must choose exactly n2n2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n2n2 nonempty piles).

Given the starting configuration, determine who will win the game.

Input

The first line contains one integer nn (2≤n≤502≤n≤50) — the number of piles. It is guaranteed that nn is an even number.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤501≤ai≤50) — the number of stones in the piles.

Output

Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).

Examples

2

8 8

Bob

4

3 1 4 1

Alice

题意：
给偶数（n）堆石子，每一步必须取n/2堆石子中的任意多个，当场上不足n/2堆石子时，当前玩家失败，问谁是最后的获胜者。
思路：
如果有任何一堆石子已经被拿空，那么只需要直接取空n/2堆石子，便可以获胜。
所以作为后手，如果能维护石子数量最小的堆数量大于N，便可以取胜，因为在这种情况下，石子数越来越少，先手总会拿空一堆。

#include<iostream>

#include<algorithm>

#include<vector>

#include<stack>

#include<queue>

#include<map>

#include<set>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<ctime>

#define fuck(x) cout<<#x<<" = "<<x<<endl;

#define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl;

#define ls (t<<1)

#define rs ((t<<1)+1)

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int maxn = ;

const int maxm = ;

const int inf = 2.1e9;

const ll Inf = ;

const int mod = ;

const double eps = 1e-;

const double pi = acos(-);

int num[maxn];

int main()

{

//    ios::sync_with_stdio(false);

//    freopen("in.txt","r",stdin);

    int n;

    scanf("%d",&n);

    for(int i=;i<=n;i++){

        scanf("%d",&num[i]);

    }

    sort(num+,num++n);

    if(num[n/+]!=num[]){printf("Alice\n");}

    else printf("Bob\n");

    return ;

}