Again Stone Game

Alice and Bob are playing a stone game. Initially there are n piles of stones and each pile contains some stone. Alice stars the game and they alternate moves. In each move, a player has to select any pile and should remove at least one and no more than half stones from that pile. So, for example if a pile contains 10 stones, then a player can take at least 1 and at most 5 stones from that pile. If a pile contains 7 stones; at most 3 stones from that pile can be removed.

Both Alice and Bob play perfectly. The player who cannot make a valid move loses. Now you are given the information of the piles and the number of stones in all the piles, you have to find the player who will win if both play optimally.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 1000). The next line contains n space separated integers ranging in [1, 10⁹]. The i^th integer in this line denotes the number of stones in the i^th pile.

Output

For each case, print the case number and the name of the player who will win the game.

Sample Input

5

1

1

3

10 11 12

5

1 2 3 4 5

2

4 9

3

1 3 9

Sample Output

Case 1: Bob

Case 2: Alice

Case 3: Alice

Case 4: Bob

Case 5: Alice

/*
* @Author: lyuc
* @Date:   2017-04-26 21:32:51
* @Last Modified by:   lyuc
* @Last Modified time: 2017-04-26 22:13:13
*/

/*题意：给你n堆石子每次你可以选择一堆，至少取一个，最多不能超过一半，比如7个的时候，你最多取3个谁不
 *     能取了就输了
 *
 *初步思路：想到了SG，但是没办法数据量太大了，只能打表找规律，找到规律，如果n为奇数sg(n)=sg(n/2)
 *    如果是偶数sg(n)=n/2
 */
#include <bits/stdc++.h>
using namespace std;
int t,n;
int num;
int sg[1005];
int hash[1005];
int res;
//sg打表
void init(){
    memset(sg,0,sizeof sg);
    for(int i=1;i<1005;i++){
        memset(hash,0,sizeof hash);
        for(int j=1;j+j<=i;j++){
            hash[sg[i-j]]=1;
        }
        for(int j=0;j<1005;j++){
            if(hash[j]==0){
                sg[i]=j;
                break;
            }
        }
    }
}
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++){
        printf("Case %d: ",ca);
        res=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d",&num);
            while(num%2!=0){
                num/=2;
            }
            res^=num;
        }
        printf(res==0?"Bob\n":"Alice\n");
    }
    return 0;
}