Alice and Bob are playing game of Misère Nim. Misère Nim is a game playing on k piles of stones, each pile containing one or more stones. The players alternate turns and in each turn a player can select one of the piles and can remove as many stones from that pile unless the pile is empty. In each turn a player must remove at least one stone from any pile. Alice starts first. The player who removes the last stone loses

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case starts with a line containing an integer k (1 ≤ k ≤ 100). The next line contains k space separated integers denoting the number of stones in each pile. The number of stones in a pile lies in the range [1, 109]

Output

For each case, print the case number and 'Alice' if Alice wins otherwise print 'Bob'.

Sample Input

3

4

2 3 4 5

5

1 1 2 4 10

1

1

Sample Output

Case 1: Bob

Case 2: Alice

Case 3: Bob

題意：

有N堆石子，每次可以從一堆中取任意個石子，至少一個，拿最後一個石子的人輸（經典Nim博弈為拿走最後一顆石子勝）

當N堆石子都為1時，奇數個石子堆，先手必敗；其他情況符合Nim博弈性質。Nim博弈的必勝態在這個遊戲中任然是必勝態，因為處於必勝態的人一定可以把最後一顆石子留給對方

#include <cstdio>
#include <cstring>
#include <algorithm>
#include<iostream>
using namespace std;
typedef long long LL;
const int maxn=120;
LL a[120];
int main(){
    int T;
    int ca=1;
    scanf("%d",&T);
    while(T--){
        int n,m;
        scanf("%d",&n);
        LL ans=0;
       
        bool flag=true;
        for(int i=0;i<n;i++){
            scanf("%lld",&a[i]);
            if(a[i]!=1)  flag=false;
            ans^=a[i];
        }
        printf("Case %d: ",ca++);
        if(flag){
            if(n%2==1)
                printf("Bob\n");
            else
                printf("Alice\n");
            continue;
        }
        if(ans==0)
            printf("Bob\n");
        else
            printf("Alice\n");
    }
    return 0;
}