Description

Background 
The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey 
around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans? 

Problem 
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.

Input

The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number. 
If no such path exist, you should output impossible on a single line.

Sample Input

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

題目給出格子的長和寬，要求以馬的走法能否走完全部格子，還要按字典序（ lexicographically）輸出！

模擬馬的走法（int step[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};）

以下是程式碼：

1. #include<stdio.h>  

2. #include<stdlib.h>  

3. #include<string.h>  

4. #include<math.h>  

5. #define max 27  

6. int map[max][max],x[max],y[max];  

7. int p,q,t=0,sign;  

8. int step[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};//模擬馬走的步子  

9. void cxbdfs(int a,int b)// 深搜  

10. {  

11.     int x1,y1,i;  

12.     if(sign)// 標記退出  

13.         return;  

14.     t++;  

15.     x[t]=a;  

16.     y[t]=b;  

17.     if(t==p*q) // 當全部走遍，標記  

18.     {  

19.         sign=1;  

20.         return;  

21.     }  

22.     map[a][b]=1;  

23.     for(i=0;i<8;i++)  

24.     {  

25.         x1=a+step[i][0];  

26.         y1=b+step[i][1];  

27.         if(x1>0 && y1>0 && x1<=q && y1<=p && map[x1][y1]==0)// 符合條件進入搜尋  

28.         {  

29.             cxbdfs(x1,y1);  

30.             t--;  

31.         }  

32.     }  

33.     map[a][b]=0;//不符合條件回溯時地圖從新標記  

34. }  

35. int main()  

36. {  

37.     int i,n,j=0;  

38.     scanf("%d",&n);  

39.     memset(map,0,sizeof(map));  

40.     while(n--)  

41.     {  

42.         j++;          

43.         sign=0;  

44.         scanf("%d%d",&p,&q);  

45.         cxbdfs(1,1);  

46.         printf("Scenario #%d:\n",j);  

47.         if(sign)  

48.         {  

49.             for(i=1;i<=p*q;i++)  

50.                 printf("%c%d",x[i]+'A'-1,y[i]);  

51.             printf("\n");  

52.         }  

53.         else  

54.             printf("impossible\n");  

55.         printf("%s",n==0?"":"\n");  

56.     }  

57. }