CodeForces - 1162E Thanos Nim （博弈论）

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

Input

2
8 8

Output

Bob

Input

4
3 1 4 1

Output

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 ;
}