Uva253cube Painting Dice Coloring

We have a painting cubes. It is supplied with three different colors:blue, red and green. Each face of the cube gets one of these colors. The cube ' s faces is numbered as in Figure 1.

Figure 1.

Since a cube has 6 faces, we can paint a face-numbered cube in different ways. When ignoring the face-numbers, the number of different paintings was much less, because a-cube can be rotated. See example below. We denote a painted cube by a string of 6 characters, where each character is a b, R, or g. The character () from the left gives the color of face i. For example, Figure 2 was a picture of Rbgggr and Figure 3 corresponds to rggbgr. Notice that both cubes is painted in the same way:by rotating it around the vertical axis by a, the one changes into The other.

Input

The input of your program was a textfile that ends with the standard end-of-file marker. Each line is a string of characters. The first 6 characters of this string is the representation of a painted cube, the remaining 6 characters give you the RE Presentation of another cube. Your program determines whether these-cubes is painted on the same, that's, whether by any combination of Rotati ONS one can be turned to the other. (Reflections is not allowed.)

Output

The output is a file of Boolean. For each line of input, output contains TRUE if the second half can is obtained from the first half by rotation a S describes above, FALSE otherwise.

Sample Input

Rbgggrrggbgrrrrbbbrrbbbrrbgrbgrrrrrg

Sample Output

TRUE

FALSE

FALSE

Enter a dice to determine whether it is equivalent.

1#include <iostream>2 3 using namespacestd;4            //list 4*6 species may5 inta[ -][6] =6 { {},7{1,2,3,4,5,6}, {1,4,2,5,3,6}, {1,5,4,3,2,6}, {1,3,5,2,4,6 },8{6,2,4,3,5,1}, {6,3,2,5,4,1}, {6,5,3,4,2,1}, {6,4,5,2,3,1 },9{2,6,3,4,1,5}, {2,4,6,1,3,5}, {2,1,4,3,6,5}, {2,3,1,6,4,5 },Ten{5,6,4,3,1,2}, {5,3,6,1,4,2}, {5,1,3,4,6,2}, {5,4,1,6,3,2 }, One{3,2,6,1,5,4}, {3,1,2,5,6,4}, {3,5,1,6,2,4}, {3,6,5,2,1,4 }, A{4,2,1,6,5,3}, {4,6,2,5,1,3}, {4,5,6,1,2,3}, {4,1,5,2,6,3 } - }; -  the intMain () - { -     intI, J, N, M, S, T; -     Chars1[ -], s2[ -]; +      while(CIN&GT;&GT;S1)//Input string S1 total 12 bits -     { +          for(i =6; I <= A; i++)//Assign the latter six bits to S2 A         { atS2[i-6] =S1[i]; -         } -          for(i =1; I <= -; i++) -         { -              for(j =0; J <=5; J + +) -             { in                  -                 if(S2[j] = = S1[a[i][j]-1]) to                 { +  -                 } the                  *                 Else $                 {Panax Notoginseng                      Break;//If it is not one by one, exit the current loop (the nearest for loop)
 -}//and I+1 continues to circulate .
 the} +             if(J = =6)//j==6 shows that this cycle J runs more than six times, and there are one by one correspondence.  A             { the                  Break; Exit Loop +             } -         } $         if(i = = -)//i==25 Description loop more than 24 times, no one by one corresponding entries found, False $         { -printf"false\n"); -         } the         Else -         {Wuyiprintf"true\n"); the         } -     } Wu     return 0; -}

Uva253cube Painting Dice Coloring