A. Shortest path of the king

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square t. As the king is not in habit of wasting his time, he wants to get from his current position s to square t in the least number of moves. Help him to do this.

In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).

Input

The first line contains the chessboard coordinates of square s, the second line — of square t.

Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8.

Output

In the first line print n — minimum number of the king's moves. Then in n lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.

L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.

Examples

input

a8h1

output

7RDRDRDRDRDRDRD

 解题思路：一个8*8的图，8个方向移动，字符转化为数字表示，bfs记录路径最短路径，到达目标点即遍历结束，记录路径，路径移动方向转变为题中的字符方向。

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;int d_x[4] = { 0, 1, -1 };int d_y[4] = { 0, 1, -1 };int e[10][10], book[10][10];int next[2][10000];int end_xx[10000],end_yy[10000];int sum;int start_x, start_y, end_x, end_y;struct node { int pre; int x, y; int step;};node qt[10000];void conversion(int x,int y){ if(x==1){ cout<<"R"; }else if(x==-1){ cout<<"L"; } if(y==-1){ cout<<"D"; }else if(y==1){ cout<<"U"; } }void bfs() { int head = 1; int tail = 1; node q; q.x = start_x; q.y = start_y; q.pre = head; q.step = 0; qt[tail] = q; qt[tail].pre = head; tail++; queue
          
            qq; qq.push( q ); while( !qq.empty() ) { node temp = qq.front(); head = temp.pre; // cout<<"head = "<<
           
            = 0; i-- ) {// cout<
            
             <<" "<
             
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