The way of the horse

Describe

On a 4x5 board, the position of the horse's starting position coordinates (longitudinal, transverse) is entered by the keyboard, and the horse can return to the total number of different methods of the initial position (the horse passes the position can not be repeated, the horse Walk "Day" word).

Input

Input file first behavior test Case number N, next n rows, two positive integers per line x, y (1<=x<=5,1<=y<=6), represents the position coordinates of the horse .

Output

The output of each test case is one row, and the output horse can return the total number of all different walks in its initial position, and if not, output "ERROR" (without double quotes).

Sample input

3
2 2
1 1
4 6

Sample output

4596
1508
ERROR

1#include <iostream>2#include <cstdio>3#include <cstring>4 using namespacestd;5 intm, N;6 intdx[]={1,1,-1,-1,2,2,-2,-2};7 intdy[]={2,-2,2,-2,1,-1,1,-1};8 intmap[5][6];9 intSX, Sy, step;Ten  One voidDfsintXinty) A { -     intxx, yy, I; -      for(i =0; I <8; i++) the     { -xx = x +Dx[i]; -yy = y +Dy[i]; -         if(xx>=0&& yy>=0&& xx<4&& yy<5&& Map[xx][yy] = =0)     +         { -MAP[XX][YY] =1; + DFS (XX,YY); AMAP[XX][YY] =0; at         } -         if(xx+1==sx&&yy+1==Sy) -step++; -     } - } -  in intMain () - { to           intN; +            while(cin>>N) -           { the                while(n--) *               { $Cin>>sx>>Sy;Panax Notoginseng                   if(sx<1|| Sx>4|| sy<1|| Sy>5) -                   { theprintf"error\n"); +                       Continue; A                   } thememset (Map,0,sizeofmap); +Step =0; -map[sx-1][sy-1] =1; $DFS (sx-1, sy-1); $printf"%d\n", step); -               } -           } the           return 0; -}

The way of the horse