Alice and Bob

Problem Description

The famous "Alice and Bob" are playing a game again. So now comes the new problem which need a person smart as you to decide the winner. The problem is as follows: They are playing on a rectangle paper, Alice and Bob take turn alternatively, for each turn, a people cut the rectangle vertically or horizontally, the result two rectangle after cut must be IDENTICAL, also the side must be integer, after the cut, one rectangle will be descarded. The first people fail to cut lose the game. Of course, Alice makes first as usual.

Input

First Line contains an integer t indicate there are t cases(1≤t≤1000) For each case: The input consists of two integers w and h(1≤w,h≤1,000,000,000), the size of rectangle.

Output

First output Case number For each case output Alice or Bob, indicate the winner.

Sample Input

2
1 2
2 2

Sample Output

Case 1: Alice
Case 2: Bob

//题意：把N*M的矩阵一直对半分，对其中的一半进行下一步操作，另一半忽略，直到不能进行下一步操作，就输了，Alice先操作。

//标程：

#include
#include
using namespace std;

int main()
{
//	freopen("a.txt","r",stdin);
    int t;
	__int64 n, m;
	cin >> t;
	for(int i = 1; i <= t; ++ i){
        scanf("%I64d%I64d",&n,&m);
		__int64 tmp = n * m;
		int cnt = 0;
		while(tmp % 2 == 0){
              cnt ++;
			  tmp /= 2;
		}
		printf("Case %d: ",i); 
		if(cnt % 2 == 1) printf("Alice\n");
		else printf("Bob\n");
	}
	return 0;
}