A children’s puzzle that was popular  years ago consisted of a × frame which contained  small

squares of equal size. A unique letter of the alphabet was printed on each small square. Since there

were only  squares within the frame, the frame also contained an empty position which was the same

size as a small square. A square could be moved into that empty position if it were immediately to the

right, to the left, above, or below the empty position. The object of the puzzle was to slide squares

into the empty position so that the frame displayed the letters in alphabetical order.

The illustration below represents a puzzle in its original configuration and in its configuration after

the following sequence of  moves:

) The square above the empty position moves.

) The square to the right of the empty position moves.

) The square to the right of the empty position moves.

) The square below the empty position moves.

) The square below the empty position moves.

) The square to the left of the empty position moves.

Write a program to display resulting frames given their initial configurations and sequences of moves.

Input

Input for your program consists of several puzzles. Each is described by its initial configuration and

the sequence of moves on the puzzle. The first  lines of each puzzle description are the starting

configuration. Subsequent lines give the sequence of moves.

The first line of the frame display corresponds to the top line of squares in the puzzle. The other

lines follow in order. The empty position in a frame is indicated by a blank. Each display line contains

exactly  characters, beginning with the character on the leftmost square (or a blank if the leftmost

square is actually the empty frame position). The display lines will correspond to a legitimate puzzle.

The sequence of moves is represented by a sequence of As, Bs, Rs, and Ls to denote which square

moves into the empty position. A denotes that the square above the empty position moves; B denotes

that the square below the empty position moves; L denotes that the square to the left of the empty

position moves; R denotes that the square to the right of the empty position moves. It is possible that

there is an illegal move, even when it is represented by one of the  move characters. If an illegal move

occurs, the puzzle is considered to have no final configuration. This sequence of moves may be spread

over several lines, but it always ends in the digit . The end of data is denoted by the character Z.

Output

Output for each puzzle begins with an appropriately labeled number (Puzzle #, Puzzle #, etc.). If

the puzzle has no final configuration, then a message to that effect should follow. Otherwise that final

configuration should be displayed.

Format each line for a final configuration so that there is a single blank character between two

adjacent letters. Treat the empty square the same as a letter. For example, if the blank is an interior

position, then it will appear as a sequence of  blanks — one to separate it from the square to the left,

one for the empty position itself, and one to separate it from the square to the right.

Separate output from different puzzle records by one blank line.

Note: The first record of the sample input corresponds to the puzzle illustrated above.

Sample Input

TRGSJ

XDOKI

M VLN

WPABE

UQHCF

ARRBBL0

ABCDE

FGHIJ

KLMNO

PQRS

TUVWX

AAA

LLLL0

ABCDE

FGHIJ

KLMNO

PQRS

TUVWX

AAAAABBRRRLL0

Z

Sample Output

Puzzle #:

T R G S J

X O K L I

M D V B N

W P A E

U Q H C F

Puzzle #:

A B C D

F G H I E

K L M N J

P Q R S O

T U V W X

Puzzle #:

This puzzle has no final configuration.

#include<stdio.h>

#include<string.h>

#include<algorithm>

using namespace std;

int main()

{

    char a[][],b[];

    int i,j,m,n,l,x,y,count=,boom;

    while(gets(a[]))      //输入  第一行的 数据

    {

        if(a[][]=='Z')   //如果 出现了Z  就代表 程序 结束

            break;

        for(i=;i<=;i++)  //如果没有结束的 话 那么就输入 剩下的  四行 数据

            gets(a[i]);

        for(i=;;i++)

        {

            scanf("%c",&b[i]);

            if(b[i]=='')

                break;

        }

        //scanf("%s",b);     //   输入指令 .

        l=strlen(b)-;    //得到指令的长度    (因为 指令  以 0 作为结束 所以...)

        for(i=;i<;i++)

        {

            for(j=;j<;j++)

            {

                if(a[i][j]==' ')

                    {

                        x=i;      //现在  得到了  空格 在 4*4 格子中的

                        y=j;

                        i=;

                        break;

                    }

            }

            if(i==)

                break;

        }

        for(i=;i<l;i++)

        {

            if(b[i]=='A')

            {

                if(x-<)

                {

                    boom=;

                    break;

                }

                swap(a[x-][y],a[x][y]);

                x--;

            }

            else

                if(b[i]=='B')

            {

                if(x+>)

                {

                    boom=;

                    break;

                }

                swap(a[x+][y],a[x][y]);

                x++;

            }

            else

                if(b[i]=='R')

            {

                if(y+>)

                {

                    boom=;

                    break;

                }

                swap(a[x][y+],a[x][y]);

                y++;

            }

            else

                if(b[i]=='L')

            {

                if(y-<)

                {

                    boom=;

                    break;

                }

                swap(a[x][y],a[x][y-]);

                y--;

            }

        }

        if(count!=)

            printf("\n");

        printf("Puzzle #%d:\n",count++);

        if(boom==)

        {

            printf("This puzzle has no final configuration.\n");

            boom=;

        }

        else

        for(i=;i<;i++)

        {

            printf("%c",a[i][]);

            for(j=;j<;j++)

            {

                printf(" %c",a[i][j]);

            }

            printf("\n");

        }

        memset(a,'',sizeof(a));

        getchar();

    }

    return ;

}