# John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 2179 Accepted Submission(s): 1175

Problem Description

Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to eat several M&Ms of the same color. Then his opponent has to make a turn. And so on.
Please note that each player has to eat at least one M&M during his turn. If John (or his brother) will eat the last M&M from the box he will be considered as a looser and he will have to buy a new candy box.

Both of players are using optimal game strategy. John starts first always. You will be given information about M&Ms and your task is to determine a winner of such a beautiful game.

Both of players are using optimal game strategy. John starts first always. You will be given information about M&Ms and your task is to determine a winner of such a beautiful game.

Input

The first line of input will contain a single integer T – the number of test cases. Next T pairs of lines will describe tests in a following format. The first line of each test will contain an integer N – the amount of different M&M
colors in a box. Next line will contain N integers Ai, separated by spaces – amount of M&Ms of i-th color.

Constraints:

1 <= T <= 474,

1 <= N <= 47,

1 <= Ai <= 4747

Constraints:

1 <= T <= 474,

1 <= N <= 47,

1 <= Ai <= 4747

Output

Output T lines each of them containing information about game winner. Print “John” if John will win the game or “Brother” in other case.

Sample Input

2 3 3 5 1 1 1

Sample Output

John Brother

**先者必败条件： 奇异局势&& 充裕堆数目>=2 或者 非奇异局势&& 孤单堆数目==n**

/*************************** # 2013-8-23 17:04:27 # Time: MS Memory: KB # Author: zyh # Status: Accepted ***************************/ #include<iostream> #include<stdio.h> #include<string.h> #include<stdlib.h> #include<algorithm> using namespace std; int main() { int n,m,t,sum,cnt1,cnt2,flag,i,a[48]; scanf("%d",&t); while(t--) { scanf("%d",&n); for(sum=cnt1=cnt2=0,i=0;i<n;i++){ scanf("%d",&a[i]); if(a[i]>1) cnt1++; //充裕堆数目 else cnt2++; //孤单堆数目 sum ^= a[i]; } if((sum==0&&cnt1>=2) || (sum && cnt2==n) )printf("Brother\n"); else printf("John\n"); } return 0; }